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Commit 4ca9df3

Browse files
新增部分哈希表题目
1 parent 21e5f8e commit 4ca9df3

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/*
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* @lc app=leetcode.cn id=451 lang=cpp
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*
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* [451] 根据字符出现频率排序
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*/
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// @lc code=start
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class Solution {
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public:
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string frequencySort(string s) {
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vector<int> counting(256);
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for (char c : s) {
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counting[c]++;
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}
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string ans;
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while (ans.size() < s.size()) {
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int max_idx = 0;
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for (size_t i = 0; i < counting.size(); i++) {
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if (counting[i] > counting[max_idx]) {
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max_idx = i;
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}
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}
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ans += string(counting[max_idx], max_idx);
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counting[max_idx] = 0;
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}
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return ans;
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}
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};
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// @lc code=end
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‎哈希表/500.键盘行.cpp‎

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/*
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* @lc app=leetcode.cn id=500 lang=cpp
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*
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* [500] 键盘行
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*/
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// @lc code=start
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class Solution {
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public:
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vector<string> findWords(vector<string>& words) {
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vector<unordered_set<char>> ss(3);
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vector<string> tmp = { "qwertyuiop", "asdfghjkl", "zxcvbnm" };
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for (size_t i = 0; i < tmp.size(); i++) {
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for (char c : tmp[i]) {
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ss[i].insert(c);
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ss[i].insert(c & '_'); // 转成大写
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}
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}
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vector<string> ans;
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for (const string& w : words) {
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int idx = -1;
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for (size_t i = 0; i < ss.size(); i++) {
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if (ss[i].find(w[0]) != ss[i].end()) {
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idx = i;
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break;
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}
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}
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bool flag = true;
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for (size_t j = 1; j < w.size(); j++) {
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if (ss[idx].find(w[j]) == ss[idx].end()) {
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flag = false;
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break;
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}
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}
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if (flag) {
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ans.push_back(w);
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}
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}
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return ans;
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}
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};
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// @lc code=end
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/*
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* @lc app=leetcode.cn id=508 lang=cpp
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*
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* [508] 出现次数最多的子树元素和
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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public:
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vector<int> findFrequentTreeSum(TreeNode* root) {
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int _ = getSum(root);
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int max_count = -1;
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for (auto& kv : memo) {
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max_count = max(max_count, kv.second);
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}
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vector<int> ans;
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for (auto& kv : memo) {
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if (kv.second == max_count) {
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ans.push_back(kv.first);
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}
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}
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return ans;
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}
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int getSum(TreeNode* root) {
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if (root == nullptr)
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{
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return 0;
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}
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int sum = root->val + getSum(root->left) + getSum(root->right);
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memo[sum]++;
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return sum;
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}
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private:
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unordered_map<int, int> memo;
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};
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// @lc code=end
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‎哈希表/525.连续数组.cpp‎

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/*
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* @lc app=leetcode.cn id=525 lang=cpp
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*
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* [525] 连续数组
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*/
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// @lc code=start
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class Solution {
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public:
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int findMaxLength(vector<int>& nums) {
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// 前缀和 + 哈希表
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// 将 0 变为 -1
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// 1. 若前缀和为0,其长度为 i + 1
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// 2. 若前缀和nums[i] == nums[j], 则 [i+1, j]的元素 01个数相等, 长度为 j - i
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if (nums.size() < 2) {
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return 0;
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}
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int ans = 0;
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unordered_map<int, int> memo;
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nums[0] = nums[0] == 0 ? -1 : nums[0];
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memo[nums[0]] = 0;
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for (int i = 1; i < nums.size(); i++) {
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nums[i] = nums[i] == 0 ? -1 : nums[i];
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nums[i] = nums[i] + nums[i-1];
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if (memo.find(nums[i]) == memo.end()) {
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memo[nums[i]] = i;
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} else {
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ans = max(ans, i - memo[nums[i]]);
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}
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if (nums[i] == 0) {
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ans = max(ans, i+1);
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}
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}
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return ans;
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// 0 0 0 1 1 0 0 1 1 1
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// -1 -2 -3 -2 -1 -2 -3 -2 -1 0
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}
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};
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// @lc code=end
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‎哈希表/554.砖墙.cpp‎

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/*
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* @lc app=leetcode.cn id=554 lang=cpp
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*
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* [554] 砖墙
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*/
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// @lc code=start
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class Solution {
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public:
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int leastBricks(vector<vector<int>>& wall) {
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// 统计边界的最大个数
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unordered_map<int, int> bound;
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for (const vector<int>& v : wall)
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{
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int x = 0;
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for (int i = 0; i < v.size() - 1; i++) {
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x += v[i];
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bound[x]++;
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}
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}
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int res = 0;
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for (auto& kv : bound) {
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res = max(res, kv.second);
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}
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return wall.size() - res;
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}
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};
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// @lc code=end
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‎哈希表/575.分糖果.cpp‎

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/*
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* @lc app=leetcode.cn id=575 lang=cpp
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*
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* [575] 分糖果
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*/
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// @lc code=start
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class Solution {
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public:
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int distributeCandies(vector<int>& candyType) {
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unordered_set<int, int> memo;
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for (int c : candyType) {
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memo[c]++;
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}
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return min(candyType.size() / 2, memo.size());
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}
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};
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// @lc code=end
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/*
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* @lc app=leetcode.cn id=594 lang=cpp
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*
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* [594] 最长和谐子序列
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*/
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// @lc code=start
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class Solution {
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public:
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int findLHS(vector<int>& nums) {
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int ans = 0;
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unordered_map<int, int> memo; // 统计个数
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for (int i = 0; i < nums.size(); i++) {
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memo[nums[i]]++;
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}
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// 子序列只能包含 a, a+1 或 a-1, a
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for (auto& kv : memo) {
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int key = kv.first;
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// key+1 不存在于 memo时,
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// 使用 memo[key+1]这种写法会初始化 memo[key+1] = 0, 影响for循环对memo的遍历
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if (memo.find(key+1) != memo.end())
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ans = max(ans, memo[key] + memo[key+1]);
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if (memo.find(key-1) != memo.end())
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ans = max(ans, memo[key] + memo[key-1]);
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}
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return ans;
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}
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};
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// @lc code=end
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/*
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* @lc app=leetcode.cn id=599 lang=cpp
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*
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* [599] 两个列表的最小索引总和
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*/
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// @lc code=start
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class Solution {
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public:
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vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
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unordered_map<string, int> map1;
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unordered_map<string, int> map2;
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for (size_t i = 0; i < list1.size(); i++) {
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map1[list1[i]] = i;
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}
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for (size_t i = 0; i < list2.size(); i++) {
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map2[list2[i]] = i;
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}
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vector<string> ans;
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int sum = INT_MAX;
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for (auto& kv : map2) {
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string k = kv.first;
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if (map1.find(k) != map1.end()) {
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if (map1[k] + map2[k] < sum) {
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ans.clear();
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ans.push_back(k);
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sum = map1[k] + map2[k];
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} else if (map1[k] + map2[k] == sum) {
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ans.push_back(k);
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}
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}
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}
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return ans;
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}
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};
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// @lc code=end
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