lc57 Insert Interval 仔细分析题目,发现我们只需要处理那些与插入interval重叠的interval即可,换句话说,那些end早于插入start以及start晚于插入end的interval都可以保留.我们只需要两个指针,i&j分别保存重叠interval中最早start和最晚end即可,然后将interval [i, j]插入即可 class Solution { public int[][] insert(int[][] intervals, int[] newInte…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 栈 日期 题目地址:https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string/ 题目描述 Given a string S of lowercase letters, a duplicate removal consists of choosing two adjac…

1047. Remove All Adjacent Duplicates In String(删除字符串中的所有相邻重复项) 链接:https://leetcode-cn.com/problems/remove-all-adjacent-duplicates-in-string/ 题目: 给出由小写字母组成的字符串 S,重复项删除操作会选择两个相邻且相同的字母,并删除它们. 在 S 上反复执行重复项删除操作,直到无法继续删除. 在完成所有重复项删除操作后返回最终的字符串.答案保证唯一. 示例:…

题目如下: Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal letters, and removing them. We repeatedly make duplicate removals on S until we no longer can. Return the final string after all such duplica…

1047. 删除字符串中的所有相邻重复项 1047. Remove All Adjacent Duplicates In String 题目描述 LeetCode1047. Remove All Adjacent Duplicates In String简单 Java 实现 import java.util.Stack; class Solution { public String removeDuplicates(String S) { if (S == null || S.length()…

题目如下: Given a string s, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them causing the left and the right side of the deleted substring to concatenate together. We repeatedly make k duplicate removals on …

1046. 最后一块石头的重量 1046. Last Stone Weight 题目描述 每日一算法2019/6/22Day 50LeetCode1046. Last Stone Weight Java 实现 and so on 参考资料 https://leetcode.com/problems/last-stone-weight/ https://leetcode-cn.com/problems/last-stone-weight/…

problem 1046. Last Stone Weight 参考 1. Leetcode_easy_1046. Last Stone Weight; 完…

56. Merge Interval 0. 参考文献 序号 文献 1 花花酱 LeetCode 56. Merge Intervals 2 [LeetCode] Merge Intervals 合并区间 1. 题目 Given a collection of intervals, merge all overlapping intervals. Example 1: Input: [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]…

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). You may assume that the intervals were initially sorted according to their start times. Example 1: Input: intervals = [[1,3],[6,9]], newInterval…