思路一、很简单,利用python的特性,将nums2的前n项,给nums1的m项之后做延伸,然后对nums1做sort()即可,代码示例为 :merge2; 思路二、从后往前遍历,取三个游标,其中两个游标分别对应nums1和nums2,因为nums1的长度不限,最终合并的数组依然在nums1中,所以设置一个新的合并之后的nums1的游标,长度设置为 m+n-1...
LeetCode 88. Merge Sorted Array 程序员木子 香港浸会大学 数据分析与人工智能硕士在读Description Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: The number of elements initialized in nums1 and nums2 are m and n respectively.You may ...
Merge Sorted Array : https://leetcode.com/problems/merge-sorted-array/ 合并两个有序数组 : https://leetcode.cn/problems/merge-sorted-array/ LeetCode 日更第143天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满
Mergenums1andnums2into a single array sorted in non-decreasing order. The final sorted array should not be returned by the function, but instead bestored inside the arraynums1. To accommodate this,nums1has a length ofm + n, where the firstmelements denote the elements that should be merge...
Leetcode No.88 Merge Sorted Array(c++实现) 1. 题目 1.1 英文题目 You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively. ...
LeetCode: 88. Merge Sorted Array 题目描述 Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of ...
Description: Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and...
链接:https://leetcode.cn/problems/merge-sorted-array/ 解题思路 有三种解题思路 最简单,代码最少 将num2拷贝到num1中,再对num1进行排序即可。 代码: funcmerge(nums1[]int,mint,nums2[]int,nint)[]int{//first cody nums2 to nums1//使用copy方法需要注意索引//_ = copy(nums1, nums2)//使用appe...
voidmerge(vector<int>&nums1,intm,vector<int>&nums2,intn) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2Case 3 nums1 = [1,2,3,0,0,0] m = 3 nums2 = [2,5,6] n = 3 99 1 2 3 4 5 6 7 8 9
voidmerge(int*nums1,int nums1Size,int m,int*nums2,int nums2Size,int n){int i1=m-1,i2=n-1;int j=m+n-1;while(i1>=0&&i2>=0)//不满足一个条件就结束{if(nums1[i1]>nums2[i2]){nums1[j--]=nums1[i1--];//相当于//nums1[j] = nums1[i1];//j--;//i1--;}else{...