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Commit d6c1a1a

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新增双指针部分题解
1 parent 2f394d4 commit d6c1a1a

23 files changed

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/*
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* @lc app=leetcode.cn id=653 lang=cpp
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*
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* [653] 两数之和 IV - 输入 BST
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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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bool findTarget(TreeNode* root, int k) {
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if (root == nullptr)
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return false;
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if (st.find(k - root->val) != st.end())
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return true;
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st.insert(root->val);
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return findTarget(root->left, k) || findTarget(root->right, k);
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}
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private:
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unordered_set<int> st;
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};
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// @lc code=end
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/*
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* @lc app=leetcode.cn id=1089 lang=cpp
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*
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* [1089] 复写零
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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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// void duplicateZeros(vector<int>& arr) {
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// vector<int> a(arr);
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// int idx = 0;
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// for (int i = 0; i < a.size() - 1; i++) {
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// if (idx >= arr.size()) break;
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// if (a[i] == 0) {
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// arr[idx++] = 0;
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// if (idx >= arr.size()) break;
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// }
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// arr[idx++] = a[i];
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// }
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// }
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void duplicateZeros(vector<int>& arr) {
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int j = 0;
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for (int i = 0; j < arr.size(); i++) {
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if ((arr[i] & 0x0f) == 0) {
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j += 2; // 0 - 9 高4位为0 不用操作
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} else {
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arr[j++] |= ((arr[i] & 0x0f) << 4); // 低四位为arr[i]原来的值
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}
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}
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for (int i = 0; i < arr.size(); i++) {
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arr[i] = (arr[i] >> 4);
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}
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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=1332 lang=cpp
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*
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* [1332] 删除回文子序列
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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 removePalindromeSub(string s) {
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// 因为只有 'a' 'b'
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// 所以最多只需要删除 2 次, 一次删除所有的 'a', 一次删除所有的 'b'
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if (s.size() == 0)
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return 0;
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int i = 0;
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int j = s.size() - 1;
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while (i < j) {
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if (s[i] != s[j]) {
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return 2;
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}
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i++;
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j--;
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}
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return 1;
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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=1385 lang=cpp
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*
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* [1385] 两个数组间的距离值
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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 findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {
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sort(arr2.begin(), arr2.end());
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int cnt = 0;
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// | a - b | <= d
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// -d <= a - b <= d
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// a - d <= b <= a + d
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for (int a : arr1) {
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auto it1 = lower_bound(arr2.begin(), arr2.end(), a - d);
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auto it2 = upper_bound(arr2.begin(), arr2.end(), a + d);
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// cout << *it1 << " " << *it2 << endl;
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if (it1 < it2) continue;
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cnt++;
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}
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return cnt;
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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=532 lang=cpp
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*
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* [532] 数组中的 k-diff 数对
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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 findPairs(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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unordered_set<int> memo;
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unordered_set<int> ans;
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for (auto c : nums) {
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if (memo.find(c - k) != memo.end())
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ans.insert(c);
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if (memo.find(c + k) != memo.end())
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ans.insert(c);
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memo.insert(c);
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}
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return ans.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 id=541 lang=cpp
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*
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* [541] Reverse String II
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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 reverseStr(string s, int k) {
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int left = 0;
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int right = 0;
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int n = s.size();
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while (left < n) {
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right = min(n - 1, left + k - 1);
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reverse(s, left, right);
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left = left + 2 * k;
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}
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return s;
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}
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void reverse(string& s, int left, int right) {
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while (left < right) {
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char c = s[left];
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s[left] = s[right];
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s[right] = c;
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left++;
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right--;
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}
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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 id=557 lang=cpp
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*
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* [557] Reverse Words in a String III
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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 reverseWords(string s) {
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// reverse(s, 0, s.size() - 1);
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int left = 0;
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int right = 0;
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while (left < s.size()) {
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while (right < s.size() && s[right] != ' ')
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right++;
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// cout << right << "/" << s.size() << " ";
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reverse(s, left, right - 1);
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// cout << "22/" << " " << endl;
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left = right = right + 1;
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}
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return s;
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}
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void reverse(string& s, int left, int right) {
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while (left < right) {
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char c = s[left];
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s[left] = s[right];
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s[right] = c;
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left++;
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right--;
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}
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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 id=567 lang=cpp
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*
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* [567] Permutation in String
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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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bool checkInclusion(string s1, string s2) {
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if (s2.size() < s1.size()) return false;
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vector<int> v1(26);
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vector<int> v2(26);
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// 滑动窗口
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int n = s1.size();
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for (int i = 0; i < n; i++) {
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v1[s1[i] - 'a']++;
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v2[s2[i] - 'a']++;
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}
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if (v1 == v2) return true;
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for (int i = n; i < s2.size(); i++) {
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v2[s2[i - n] - 'a']--;
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v2[s2[i] - 'a']++;
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if (v1 == v2) return true;
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}
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return false;
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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=581 lang=cpp
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*
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* [581] 最短无序连续子数组
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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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// TODO: 未完成
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int findUnsortedSubarray(vector<int>& nums) {
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int left = 0;
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int right = 0;
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for (int i = 1; i < nums.size(); i++) {
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if (nums[i] < nums[i-1]) {
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left = i - 1;
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right = i;
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break;
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}
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}
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if (left == 0 && right == 0)
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return 0;
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cout << left << " " << right << endl;
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for (int i = right; i < nums.size(); i++) {
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cout << " i=" << i << endl;;
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if (nums[i] < nums[i - 1]) {
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// 处理重复
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while (i < nums.size() - 1 && nums[i + 1] == nums[i])
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{
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i++;
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}
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right = i;
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}
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}
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return right - left + 1;
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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=611 lang=cpp
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*
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* [611] 有效三角形的个数
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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 triangleNumber(vector<int>& nums) {
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int ans = 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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for (int j = i + 1; j < nums.size(); j++) {
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// 二分法搜索
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int left = j + 1;
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int right = nums.size() - 1;
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int b = j;
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while (left <= right) {
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int mid = left + (right - left) / 2;
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if (nums[mid] < nums[i] + nums[j]) {
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left = mid + 1;
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b = mid;
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} else {
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right = mid - 1;
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}
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}
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// cout << left << " " << right << " " << j << endl;
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ans += (b - j);
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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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