题目:
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.
解题思路1:设置一个数组C,依次取A的元素i与B的元素j比较,如果B[j]小,则将B[j]放入C,再取B的后一个元素与A[i]比较;反之,若A[i]更小,则将A[i]放入C,取A的下一个元素与B[j]比较。比较完成后,将C复制到数组A。该解法的时间复杂度为O(m+n),空间复杂度为O(m+n)
代码1:
class Solution { public: void merge(int A[], int m, int B[], int n) { int C[m+n]; int i=0,j=0,k=0; while(i<m&&j<n){ if(A[i]<B[j]){ C[k++]=A[i++]; }else{ C[k++]=B[j++]; } } while(i<m){ C[k++]=A[i++]; } while(j<n){ C[k++]=B[j++]; } for(i=0;i<m+n;i++){ A[i]=C[i]; } } };
代码2:
class Solution { public: void merge(int A[], int m, int B[], int n) { int mn=m+n; int ptr_a=m-1,ptr_b=n-1,ptr_mn=m+n-1; while(ptr_a>=0&&ptr_b>=0){ A[ptr_mn--]=A[ptr_a]>B[ptr_b]?A[ptr_a--]:B[ptr_b--]; } while(ptr_b>=0){ A[ptr_mn--]=B[ptr_b--]; } } };
【Leetcode】Merge Sorted Array,布布扣,bubuko.com
原文:http://blog.csdn.net/ussam/article/details/22087035