面试题：两个无序数组合并成一个有序数组

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昨天面试，问了一个简单的算法题，当时快速排序的代码有点忘了，还想了一会儿。最后写的估计也还有点问题，再写一遍。

思路：先分别对两个数组进行排序（可以用快速排序），然后两路归并。

#include <iostream>using namespace std;class Solution {public:int *sort(int *a,int lenA,int *b,int lenB){fastSort(a,0,lenA);fastSort(b,0,lenB);return merge(a,lenA,b,lenB);}private://快速排序void fastSort(int *a,int start,int end){if(a==NULL || end-start<=1 || start<0)return;int pivotPos = start;int pivot = a[start];int temp;for(int i=start+1;i<end;++i){if(a[i]<pivot){if(++pivotPos!=i){temp = a[i];a[i] = a[pivotPos];a[pivotPos] = temp;}}}a[start] = a[pivotPos];a[pivotPos] = pivot;fastSort(a,start,pivotPos-1);fastSort(a,pivotPos+1,end);}//两路归并int *merge(int *a,int lenA,int *b,int lenB){if(a==NULL || lenA<=0)return b;if(b==NULL || lenB<=0)return a;int *arry = new int[lenA+lenB];if(arry==NULL){cerr << "内存分配失败" << endl;exit(1);}int posA = 0, posB = 0 ,pos = 0;while(posA<lenA && posB<lenB){if(a[posA]<b[posB])arry[pos++] = a[posA++];elsearry[pos++] = b[posB++];}while(posA<lenA)arry[pos++] = a[posA++];while(posB<lenB)arry[pos++] = b[posB++];return arry;}};void main(){int len = 5;int *a = new int[len];int *b = new int[len];for(int i=0;i<len;++i){a[i] = rand();b[i] = rand();}for(int i=0;i<len;++i)cout << a[i] << " ";cout << endl;for(int i=0;i<len;++i)cout << b[i] << " ";cout << "\n排序后：" << endl;int *arry = Solution().sort(a,len,b,len);for(int i=0;i<2*len;++i)cout << arry[i] << " ";cout << endl;}

假设两个数组长度相等，均为N，则算法的时间复杂度是O(2NlogN)。

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