[LeetCode53]Merge Sorted Array

最新推荐文章于 2025-08-08 11:49:43 发布

Mavs

最新推荐文章于 2025-08-08 11:49:43 发布

阅读量645

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode LeetCode Questions 文章标签： leetcode array mergesort

本文链接：https://blog.youkuaiyun.com/sbitswc/article/details/33497865

Leetcode 同时被 2 个专栏收录

287 篇文章

订阅专栏

LeetCode Questions

178 篇文章

订阅专栏

本文介绍了一个经典的算法问题：如何将两个已排序的整数数组合并为一个有序数组。该问题通常出现在《Cracking the Code Interview》等编程面试指南中。文章提供了一种从后向前比较并合并数组的方法，并给出了 Java 和 C++ 的实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 are m andn respectively.

Analysis:

The classic problem. (can be found in the book "Cracking the Code Interview").
Part of the merge sort, merge the arrays from the back by comparing the elements.

Java

public void merge(int A[], int m, int B[], int n) {
        int count = m+n-1;
        m--;n--;
        while(m>=0 && n>=0){
        	A[count--] = A[m]>B[n]? A[m--]:B[n--];
        }
        while(n>=0)
        	A[count--] = B[n--];
    }

c++

void merge(int A[], int m, int B[], int n) {
        int count = m+n-1;
        m--;n--;
        while(m>=0 && n>=0){
        	A[count--] = A[m]>B[n]? A[m--]:B[n--];
        }
        while(n>=0)
        	A[count--] = B[n--];
    }