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Commit 307b2bc

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cpp
Change-Id: I90b2c24989f146480df491d55865d71b3b762505
1 parent 54454b7 commit 307b2bc

8 files changed

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‎cpp/leetcode/1.两数之和.cpp

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/*
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* @lc app=leetcode.cn id=1 lang=cpp
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*
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* [1] 两数之和
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*
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* https://leetcode.cn/problems/two-sum/description/
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*
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* algorithms
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* Easy (52.84%)
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* Likes: 16011
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* Dislikes: 0
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* Total Accepted: 4.1M
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* Total Submissions: 7.7M
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* Testcase Example: '[2,7,11,15]\n9'
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*
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* 给定一个整数数组 nums 和一个整数目标值 target,请你在该数组中找出 和为目标值
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* target 的那 两个 整数,并返回它们的数组下标。
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*
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* 你可以假设每种输入只会对应一个答案。但是,数组中同一个元素在答案里不能重复出现。
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*
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* 你可以按任意顺序返回答案。
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*
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*
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*
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* 示例 1:
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*
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*
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* 输入:nums = [2,7,11,15], target = 9
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* 输出:[0,1]
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* 解释:因为 nums[0] + nums[1] == 9 ,返回 [0, 1] 。
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*
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*
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* 示例 2:
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*
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*
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* 输入:nums = [3,2,4], target = 6
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* 输出:[1,2]
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*
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*
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* 示例 3:
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*
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*
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* 输入:nums = [3,3], target = 6
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* 输出:[0,1]
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*
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*
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*
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*
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* 提示:
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*
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*
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* 2 <= nums.length <= 10^4
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* -10^9 <= nums[i] <= 10^9
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* -10^9 <= target <= 10^9
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* 只会存在一个有效答案
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*
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*
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*
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*
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* 进阶:你可以想出一个时间复杂度小于 O(n^2) 的算法吗?
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*
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*/
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// @lc code=start
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#include <iostream>
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#include <unordered_map>
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#include <vector>
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class Solution {
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public:
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std::vector<int> twoSum(std::vector<int> &nums, int target) {
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std::unordered_map<int, int> store;
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for (int i = 0; i < nums.size(); i++) {
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int rest = target - nums[i];
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auto it = store.find(rest);
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if (it != store.end()) {
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return {it->second, i};
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}
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store[nums[i]] = i;
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}
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return {};
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}
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};
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// @lc code=end
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int main(int argc, const char **argv) {
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Solution s;
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std::vector<int> arr = {2, 3, 4, 5};
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auto ret = s.twoSum(arr, 6);
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std::cout << "ret: ";
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for (size_t i = 0; i < ret.size(); i++) {
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std::cout << ret[i] << ", ";
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}
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std::cout << std::endl;
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return 0;
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}
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/*
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* @lc app=leetcode.cn id=11 lang=cpp
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*
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* [11] 盛最多水的容器
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*
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* https://leetcode-cn.com/problems/container-with-most-water/description/
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*
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* algorithms
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* Medium (63.82%)
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* Likes: 2531
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* Dislikes: 0
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* Total Accepted: 454.7K
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* Total Submissions: 712.4K
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* Testcase Example: '[1,8,6,2,5,4,8,3,7]'
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*
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* 给你 n 个非负整数 a1,a2,...,an,每个数代表坐标中的一个点 (i, ai)
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* 。在坐标内画 n 条垂直线,垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0)
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* 。找出其中的两条线,使得它们与 x 轴共同构成的容器可以容纳最多的水。
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*
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* 说明:你不能倾斜容器。
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*
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*
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*
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* 示例 1:
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*
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*
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*
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*
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* 输入:[1,8,6,2,5,4,8,3,7]
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* 输出:49
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* 解释:图中垂直线代表输入数组
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* [1,8,6,2,5,4,8,3,7]。在此情况下,容器能够容纳水(表示为蓝色部分)的最大值为 49。
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*
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* 示例 2:
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*
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*
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* 输入:height = [1,1]
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* 输出:1
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*
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*
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* 示例 3:
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*
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*
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* 输入:height = [4,3,2,1,4]
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* 输出:16
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*
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*
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* 示例 4:
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*
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*
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* 输入:height = [1,2,1]
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* 输出:2
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*
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*
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*
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*
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* 提示:
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*
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*
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* n = height.length
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* 2
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* 0
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*
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*
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*/
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// @lc code=start
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#include <algorithm>
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#include <iostream>
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#include <vector>
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class Solution {
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public:
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int maxArea(std::vector<int> &height) {
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int l = 0, r = height.size() - 1, ma = 0;
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while (l < r) {
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if (height.at(l) < height.at(r)) {
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ma = std::max(ma, (r - l) * height.at(l));
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l++;
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} else {
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ma = std::max(ma, (r - l) * height.at(r));
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r--;
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}
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}
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return ma;
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}
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};
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// @lc code=end
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void print_ret(int ret) {
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std::cout << "ret: ";
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std::cout << ret << " ";
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std::cout << std::endl;
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}
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int main(int argc, const char **argv) {
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Solution s;
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std::vector<int> arr = {1,8,6,2,5,4,8,3,7};
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auto ret = s.maxArea(arr);
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print_ret(ret);
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return 0;
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}

‎cpp/leetcode/15.三数之和.cpp

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/*
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* @lc app=leetcode.cn id=15 lang=cpp
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*
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* [15] 三数之和
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*
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* https://leetcode-cn.com/problems/3sum/description/
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*
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* algorithms
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* Medium (32.32%)
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* Likes: 3410
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* Dislikes: 0
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* Total Accepted: 532.8K
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* Total Submissions: 1.6M
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* Testcase Example: '[-1,0,1,2,-1,-4]'
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*
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* 给你一个包含 n 个整数的数组 nums,判断 nums 中是否存在三个元素 a,b,c
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* ,使得 a + b + c = 0 ?请你找出所有和为 0 且不重复的三元组。
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*
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* 注意:答案中不可以包含重复的三元组。
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*
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*
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*
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* 示例 1:
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*
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*
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* 输入:nums = [-1,0,1,2,-1,-4]
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* 输出:[[-1,-1,2],[-1,0,1]]
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*
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*
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* 示例 2:
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*
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*
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* 输入:nums = []
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* 输出:[]
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*
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*
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* 示例 3:
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*
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*
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* 输入:nums = [0]
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* 输出:[]
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*
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*
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*
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*
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* 提示:
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*
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*
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* 0
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* -10^5
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*
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*
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*/
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// @lc code=start
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#include <algorithm>
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#include <iostream>
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#include <ostream>
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#include <vector>
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class Solution {
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public:
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std::vector<std::vector<int>> threeSum(std::vector<int> &nums) {
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std::vector<std::vector<int>> ret;
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if (nums.size() < 3) {
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return ret;
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}
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std::sort(nums.begin(), nums.end());
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if (nums.at(nums.size() - 1) < 0) {
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return ret;
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}
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for (size_t i = 0; i < nums.size() - 2; i++) {
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if (nums.at(0) > 0) {
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break;
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}
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if (i > 0 && nums.at(i) == nums.at(i - 1)) {
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continue;
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}
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int rest = 0 - nums.at(i);
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int l = i + 1, r = nums.size() - 1;
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while (l < r) {
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int tmp = nums.at(l) + nums.at(r);
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if (tmp == rest) {
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ret.push_back({nums.at(i), nums.at(l), nums.at(r)});
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while (l < r && nums.at(l) == nums.at(l + 1)) {
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l++;
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}
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while (l < r && nums.at(r) == nums.at(r - 1)) {
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r--;
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}
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l++;
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r--;
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} else if (tmp < rest) {
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l++;
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} else {
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r--;
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}
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}
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}
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return ret;
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}
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};
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// @lc code=end
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void print_ret(const std::vector<std::vector<int>> &ret) {
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std::cout << "ret: [" << std::endl;
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for (size_t i = 0; i < ret.size(); i++) {
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std::cout << "[";
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for (size_t j = 0; j < ret.at(i).size(); j++) {
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std::cout << ret.at(i).at(j) << ",";
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}
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std::cout << "]" << std::endl;
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}
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std::cout << "]" << std::endl;
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}
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int main(int argc, const char **argv) {
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Solution s;
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std::vector<int> arr = {-1, 0, 1, 2, -1, -4};
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auto ret = s.threeSum(arr);
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print_ret(ret);
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return 0;
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}
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/*
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* @lc app=leetcode.cn id=16 lang=cpp
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*
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* [16] 最接近的三数之和
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*
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* https://leetcode-cn.com/problems/3sum-closest/description/
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*
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* algorithms
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* Medium (40.62%)
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* Likes: 187
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* Dislikes: 0
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* Total Accepted: 26.7K
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* Total Submissions: 65.8K
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* Testcase Example: '[-1,2,1,-4]\n1'
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*
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* 给定一个包括 n 个整数的数组 nums 和
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* 一个目标值 target。找出 nums 中的三个整数,使得它们的和与 target
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* 最接近。返回这三个数的和。假定每组输入只存在唯一答案。
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*
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* 例如,给定数组 nums = [-1,2,1,-4], 和 target = 1.
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*
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* 与 target 最接近的三个数的和为 2. (-1 + 2 + 1 = 2).
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*
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*
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*/
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// @lc code=start
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#include <algorithm>
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#include <iostream>
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#include <vector>
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class Solution {
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public:
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int threeSumClosest(std::vector<int> &nums, int target) {
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int minDiff = INT_MAX;
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int closest = 0;
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sort(nums.begin(), nums.end());
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for (int i = 0; i < nums.size(); i++) {
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int j = i + 1, k = nums.size() - 1;
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while (j < k) {
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int threeSum = nums[i] + nums[j] + nums[k];
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int diff = abs(target - threeSum);
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if (diff == 0) {
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return 0;
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}
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if (diff < minDiff) {
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minDiff = diff;
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closest = threeSum;
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}
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if (threeSum > target) {
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k--;
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} else {
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j++;
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}
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}
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}
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return closest;
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}
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};
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// @lc code=end
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int main(int argc, const char **argv) {
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Solution s;
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std::vector<int> nums = {2,3,4,5};
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auto str = s.threeSumClosest(nums, 17);
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std::cout << str << std::endl;
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return 0;
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}

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