【Leetcode】Merge Sorted Array

最新推荐文章于 2025-09-08 16:25:29 发布

转载最新推荐文章于 2025-09-08 16:25:29 发布 · 136 阅读

文章标签：

#leetcode #java #array #merge

本文提供了一个Java实现，展示了如何将两个已排序的整数数组合并为一个有序数组。

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.

Java:

自己模仿写一下：

http://blog.youkuaiyun.com/linhuanmars/article/details/19712333

public class Solution {
    public void merge(int A[], int m, int B[], int n) {
        int k=m+n-1;
        int i=m-1;
        int j=n-1;
        while(i>=0&&j>=0)
        {
            if(A[i]>=B[j])
            {
            A[k]=A[i];
            k--;
            i--;
                
            }
            else
            {
                A[k]=B[j];
            k--;
            j--;
                
            }
        }
        
            while(j>=0){
                A[k]=B[j];
                j--;
                k--;
            }
    }
}

Reference:

1.http://blog.youkuaiyun.com/lilong_dream/article/details/22677537

public class Solution {  
    public void merge(int A[], int m, int B[], int n) {  
        int idx = m + n - 1;  
        int i = m - 1;  
        int j = n - 1;  
  
        while (i >= 0 && j >= 0) {  
            if (A[i] >= B[j]) {  
                A[idx--] = A[i--];  
            } else {  
                A[idx--] = B[j--];  
            }  
        }  
  
        if (i == -1) {  
            while (j >= 0) {  
                A[j] = B[j];  
                j--;  
            }  
        }  
    }  
}

2.http://gongxuns.blogspot.com/2012/12/leetcode-merge-sorted-array.html

public class Solution {
    public void merge(int A[], int m, int B[], int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int i = m-1,
            j = n-1,
            k = m+n-1;        
        while(k>=0){
            if(j<0 || (i>=0 && A[i]>B[j]))
                A[k--]=A[i--];
            else
                A[k--]=B[j--];
        }
    }
}