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;…

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…

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…

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: 1.将区间按照起始位置从小到大排序: 2.一次遍历,如果发现当前区间起始小于上一个区间结束,则进行合并: 解题步骤: 1.因为需要比较结构体,所以编写比较函数< 2.新建一个结果数组,保存合并后的结果: 3.…

这道题是LeetCode里的第88道题. 题目描述: 给定两个有序整数数组 nums1 和 nums2,将 nums2 合并到 nums1 中,使得 num1 成为一个有序数组. 说明: 初始化 nums1 和 nums2 的元素数量分别为 m 和 n. 你可以假设 nums1 有足够的空间(空间大小大于或等于 m + n)来保存 nums2 中的元素. 示例: 输入: nums1 = [1,2,3,0,0,0], m = 3 nums2 = [2,5,6], n = 3 输出: [1,2,2,…

题目描述 Given two sorted integer arrays A and B, merge B into A as one sorted array. Note: You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B…

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 题目意思很简单,就是合并两个有序链表,头需要是原先的两个链表中的一个.这道题在本科生的数据结构书上就有讲,原理就就是:两个链表A,B分别逐个遍历,判断两个元素的大小,取小的作为新链表的下一个节点.让小学生做,他们也能…

Merge k Sorted Lists Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.   采用优先队列priority_queue 把ListNode放入优先队列中,弹出最小指后,如果该ListNode有下一个元素,则把下一个元素放入到队列中     /** * Definition for singly-linked list. * stru…

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 思路:使用伪头部 class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { ListNode fakehead();…