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Commit ef93bdc

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Change-Id: I57ba633bc949f310ab9a1adb764e6ef1acfe65ca
1 parent 307b2bc commit ef93bdc

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/*
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* @lc app=leetcode.cn id=19 lang=cpp
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*
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* [19] 删除链表的倒数第 N 个结点
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*
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* https://leetcode.cn/problems/remove-nth-node-from-end-of-list/description/
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*
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* algorithms
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* Medium (45.01%)
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* Likes: 2376
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* Dislikes: 0
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* Total Accepted: 1M
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* Total Submissions: 2.3M
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* Testcase Example: '[1,2,3,4,5]\n2'
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*
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* 给你一个链表,删除链表的倒数第 n 个结点,并且返回链表的头结点。
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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:head = [1,2,3,4,5], n = 2
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* 输出:[1,2,3,5]
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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:head = [1], n = 1
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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:head = [1,2], n = 1
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* 输出:[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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* 链表中结点的数目为 sz
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* 1 <= sz <= 30
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* 0 <= Node.val <= 100
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* 1 <= n <= sz
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*
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*
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*
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*
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* 进阶:你能尝试使用一趟扫描实现吗?
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*
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*/
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// Definition for singly-linked list.
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struct ListNode {
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int val;
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ListNode *next;
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ListNode() : val(0), next(nullptr) {}
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ListNode(int x) : val(x), next(nullptr) {}
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ListNode(int x, ListNode *next) : val(x), next(next) {}
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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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ListNode *removeNthFromEnd(ListNode *head, int n) {
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ListNode dummy;
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dummy.next = head;
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ListNode *pre = &dummy;
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ListNode *cur = head;
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for (int i = 0; i < n; i++) {
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if (cur == nullptr) {
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return head;
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}
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cur = cur->next;
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}
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while (cur != nullptr) {
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pre = pre->next;
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cur = cur->next;
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}
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ListNode *tmp = pre->next;
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pre->next = tmp->next;
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tmp->next = nullptr;
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return dummy.next;
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}
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};
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// @lc code=end

‎cpp/leetcode/22.括号生成.cpp

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/*
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* @lc app=leetcode.cn id=22 lang=cpp
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*
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* [22] 括号生成
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*
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* https://leetcode.cn/problems/generate-parentheses/description/
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*
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* algorithms
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* Medium (77.56%)
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* Likes: 3031
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* Dislikes: 0
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* Total Accepted: 639.4K
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* Total Submissions: 824.2K
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* Testcase Example: '3'
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*
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* 数字 n 代表生成括号的对数,请你设计一个函数,用于能够生成所有可能的并且
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* 有效的 括号组合。
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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:n = 3
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* 输出:["((()))","(()())","(())()","()(())","()()()"]
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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:n = 1
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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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* 1 <= n <= 8
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*
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*
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*/
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#include <string>
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#include <vector>
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// @lc code=start
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class Solution {
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private:
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void helper(int left, int right, std::string &cur,
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std::vector<std::string> &result) {
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if (left > right) {
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return;
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}
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if (left == 0 && right == 0) {
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result.push_back(cur);
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return;
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}
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if (left > 0) {
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cur.push_back('(');
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helper(left - 1, right, cur, result);
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cur.pop_back();
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}
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if (right > 0) {
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cur.push_back(')');
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helper(left, right - 1, cur, result);
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cur.pop_back();
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}
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}
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public:
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std::vector<std::string> generateParenthesis(int n) {
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std::vector<std::string> result;
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std::string cur;
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helper(n, n, cur, result);
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return result;
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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=23 lang=cpp
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*
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* [23] 合并K个升序链表
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*
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* https://leetcode-cn.com/problems/merge-k-sorted-lists/description/
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*
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* algorithms
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* Hard (55.28%)
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* Likes: 1342
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* Dislikes: 0
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* Total Accepted: 268.7K
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* Total Submissions: 486K
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* Testcase Example: '[[1,4,5],[1,3,4],[2,6]]'
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*
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* 给你一个链表数组,每个链表都已经按升序排列。
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*
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* 请你将所有链表合并到一个升序链表中,返回合并后的链表。
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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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* 输入:lists = [[1,4,5],[1,3,4],[2,6]]
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* 输出:[1,1,2,3,4,4,5,6]
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* 解释:链表数组如下:
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* [
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* ⁠ 1->4->5,
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* ⁠ 1->3->4,
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* ⁠ 2->6
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* ]
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* 将它们合并到一个有序链表中得到。
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* 1->1->2->3->4->4->5->6
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*
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*
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* 示例 2:
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*
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* 输入:lists = []
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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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* 输入:lists = [[]]
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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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* k == lists.length
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* 0 <= k <= 10^4
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* 0 <= lists[i].length <= 500
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* -10^4 <= lists[i][j] <= 10^4
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* lists[i] 按 升序 排列
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* lists[i].length 的总和不超过 10^4
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*
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*
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*/
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// Definition for singly-linked list.
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#include <queue>
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#include <vector>
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struct ListNode {
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int val;
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ListNode *next;
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ListNode() : val(0), next(nullptr) {}
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ListNode(int x) : val(x), next(nullptr) {}
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ListNode(int x, ListNode *next) : val(x), next(next) {}
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};
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// @lc code=start
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class Solution {
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private:
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struct Element {
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ListNode *ptr;
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bool operator<(const Element &tar) const {
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return ptr->val > tar.ptr->val; // 小顶堆,升序排序用
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}
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};
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public:
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ListNode *mergeKLists(std::vector<ListNode *> &lists) {
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ListNode dummy;
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ListNode *cur = &dummy;
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std::priority_queue<Element> queue;
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for (ListNode *node : lists) {
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if (node != nullptr) {
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queue.push({node});
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}
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}
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while (!queue.empty()) {
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auto el = queue.top();
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queue.pop();
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cur->next = el.ptr;
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cur = cur->next;
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if (el.ptr->next != nullptr) {
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queue.push({el.ptr->next});
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}
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}
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return dummy.next;
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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=25 lang=cpp
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*
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* [25] K 个一组翻转链表
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*
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* https://leetcode.cn/problems/reverse-nodes-in-k-group/description/
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*
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* algorithms
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* Hard (67.74%)
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* Likes: 1907
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* Dislikes: 0
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* Total Accepted: 430.9K
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* Total Submissions: 636.1K
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* Testcase Example: '[1,2,3,4,5]\n2'
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*
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* 给你链表的头节点 head ,每 k 个节点一组进行翻转,请你返回修改后的链表。
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*
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* k
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* 是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍,那么请将最后剩余的节点保持原有顺序。
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*
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* 你不能只是单纯的改变节点内部的值,而是需要实际进行节点交换。
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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:head = [1,2,3,4,5], k = 2
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* 输出:[2,1,4,3,5]
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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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*
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*
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* 输入:head = [1,2,3,4,5], k = 3
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* 输出:[3,2,1,4,5]
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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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* 链表中的节点数目为 n
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* 1 <= k <= n <= 5000
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* 0 <= Node.val <= 1000
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*
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*
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*
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*
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* 进阶:你可以设计一个只用 O(1) 额外内存空间的算法解决此问题吗?
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*
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*
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*
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*
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*/
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// Definition for singly-linked list.
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#include <sys/_types/_wint_t.h>
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struct ListNode {
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int val;
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ListNode *next;
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ListNode() : val(0), next(nullptr) {}
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ListNode(int x) : val(x), next(nullptr) {}
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ListNode(int x, ListNode *next) : val(x), next(next) {}
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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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ListNode *reverseRange(ListNode *start, ListNode *end) {
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ListNode *cur = start->next, *next;
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while (cur->next != nullptr && cur->next != end) {
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next = cur->next; // s sn --- c n nn --- e => s n sn --- c nn --- e
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cur->next = next->next;
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next->next = start->next;
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start->next = next;
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}
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return cur; // 下一个start
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}
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ListNode *reverseKGroup(ListNode *head, int k) {
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ListNode dummy;
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dummy.next = head;
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int i = 0;
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ListNode *start = &dummy, *end = dummy.next;
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while (end != nullptr) {
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end = end->next;
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i++;
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if (i >= k) {
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start = reverseRange(start, end);
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i = 0;
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}
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}
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return dummy.next;
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}
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};
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// @lc code=end

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