Given a collection of intervals, merge all overlapping intervals. For example,Given [1,3],[2,6],[8,10],[15,18],return [1,6],[8,10],[15,18]. 思路:开始想用线段树,后来想想这个不是动态变化的没必要. 按区间的第一个值从小到大排序,然后跳过后面被覆盖的区间来找. sort折腾了好久,开始搞不定只好自己写了一个归并排序.现在代码里的sort是可用的. class…

题目链接Merge Intervals /** * Definition for an interval. * public class Interval { * int start; * int end; * Interval() { start = 0; end = 0; } * Interval(int s, int e) { start = s; end = e; } * } * 题目:LeetCode 第56题 Merge Intervals 区间合并给定一个区间的集合,将相邻区间之间…

Merge Intervals Given a collection of intervals, merge all overlapping intervals. For example,Given [1,3],[2,6],[8,10],[15,18],return [1,6],[8,10],[15,18].     先排序,然后循环合并   /** * Definition for an interval. * struct Interval { * int start; * int end;…

Merge Intervals Given a collection of intervals, merge all overlapping intervals. For example,Given [1,3],[2,6],[8,10],[15,18],return [1,6],[8,10],[15,18]. 解题思路: 根据start对区间进行排序,然后依次遍历进行区间合并. /** * Definition for an interval. * struct Interval { * int…

1. 原题链接 https://leetcode.com/problems/merge-intervals/description/ 2. 题目要求 给定一个Interval对象集合,然后对重叠的区域进行合并.Interval定义如下 例如下图中,[1, 3] 和 [2, 6]是有重叠部分的,可以合并成[1, 6] 3. 解题思路 先取第一个interval对象的 start 和 end 的值 ,然后对这个集合进行遍历.比较当前遍历对象的start是否比前一个对象的end小,小的话则说明二者存在…

Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 总结:处理整数溢出的方法 ①用数据类型转换long  或 long long ②在每次循环时先保存下数字变化之前的值,处理后单步恢复看是否相等 (比③好) ③整体恢复,看数字是否相等. 思路:注意30000这样以0结尾的数字,注意越界时返回0. 我检查越界是通过把翻转的数字再翻转回去,看是否相等. int rever…

Write an algorithm to determine if a number is "happy". A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the…

Given a singly linked list L: L0→L1→…→Ln-1→Ln,reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… You must do this in-place without altering the nodes' values. For example,Given {1,2,3,4}, reorder it to {1,4,2,3}. 思路: 先把链表分成两节,后半部分翻转,然后前后交叉连接. 大神的代码比我的简洁,注意分两节时用快…

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = ["leet", "code"]. Return true because &…

链表的归并排序 超时的代码 class Solution: def merge(self, head1, head2): if head1 == None: return head2 if head2 == None: return head1 # head1 and head2 point to the same link list if head1 == head2: return head1 head = None tail = None # the small pointer point…