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第十二题 Merge Sorted Array

时间:2017-04-26 20:28:24      阅读:140      评论:0      收藏:0      [点我收藏+]

标签:equal   不能   element   family   lin   --   osb   color   nal   

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 and nrespectively.

Solution1:新建ArrayList暂存merged list,后放回A

public class Solution {
    public void merge(int A[], int m, int B[], int n) {
        int length = m+n;
        ArrayList<Integer> merge = new ArrayList<Integer>();
        int posA =0, posB=0, i=0;
        while( i<length){
            if(posB>=n){merge.add(A[posA]); posA++; i++;}
            else if(posA>=m){ merge.add(B[posB]); posB++; i++;}
            else if(A[posA]<=B[posB]){merge.add(A[posA]); posA++; i++;}
            else{merge.add(B[posB]); posB++; i++;}
        }
        for(i=0; i<length;i++){
            A[i] = merge.get(i);
        }
    }
}
Solution2:可不能够不额外开辟空间?利用倒序。

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

Solution1:再简单点?

public class Solution {
    public void merge(int A[], int m, int B[], int n) {
        
        for(int i=m+n-1; i>=0; i--) A[i]=((m>0)&&(n==0 || A[m-1]>=B[n-1]))?A[--m]:B[--n];
        
    }
}


第十二题 Merge Sorted Array

标签:equal   不能   element   family   lin   --   osb   color   nal   

原文地址:http://www.cnblogs.com/mfmdaoyou/p/6770427.html

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