https://leetcode-cn.com/problems/merge-sorted-array/ 解法1 时间复杂度:O(n) 空间复杂度:O(n) 思路:额外使用数组来保存合并过程中的数据,比较两数组对应元素的大小,将较小的元素保存在新的数组中,最后需要将两数组中没有比较的元素添加到新数组的末尾。 voidmerge(vector<int>& nums1,intm, vector<int...
public: void merge(int A[], int m, int B[], int n) { int k = m + n; while (k-- > 0) A[k] = (n == 0 || (m > 0 && A[m-1] > B[n-1])) ? A[--m] : B[--n]; } }; 可读性较好: class Solution { public: void merge(int A[], int m, int B[], int...
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
void merge(vector<int> &nums1, int m, vector<int> &nums2, int n) { vector<int> a(m,0); int i = 0, j = 0; int index = 0; for (int k = 0; k < m; k++) { a[k] = nums1[k]; } while (i < m && j < n) { if (a[i] < nums2[j]) { nums1[index++] =...
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/merge-sorted-array著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 代码语言:javascript 代码运行次数:0 复制 Cloud Studio代码运行 示例1: 输入:nums1=[1,2,3,0,0,0],m=3,nums2=[2,5,6],n=3输出:[1,2,2,3,5...
LeetCode: 88. Merge Sorted Array 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 ...
Mergenums1 and nums2 into a single array sorted innon-decreasing order. The final sorted array should not be returned by the function, but instead be _stored inside the array _nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements...
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(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
package Array; /** * 88\. Merge Sorted Array(合并两个有序数组) * 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。 */ public class Solution88 { public static void main(String[] args) { ...