求两个排序数组的交集

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题目：有两个数组a{1,5,8,10,14,15,17,18,20,22,24,25,28}和b{2,4,6,8,10,12}，

如何求出他们之间的交集？要求效率越高越好，数组都是从小到大排序好的。

此题解法的详细说明可以参考侯捷的《STL源码剖析》书中第333页。此解法与论坛上说给出的解法相同。

C++代码实现如下：

#include <iostream>using namespace std;void arrayJiao(int ar[], int a_len, int br[], int b_len, int tmp[]){int i = 0, j = 0, k = 0;while(i < a_len && j < b_len){if(ar[i] < br[j])i++;if(ar[i] > br[j])j++;if(ar[i] == br[j]){tmp[k++] = ar[i];i++;j++;}}for(i = 0; i < k; ++i){cout<<tmp[i]<<" ";}cout<<endl;}int main(){int a[] = {1,2,3,4,5};int b[] = {1,2,3,6,9,0};int tem[10];arrayJiao(a,5,b,6,tem);return 0;}

或者为：

#include <iostream>using namespace std;void Array_Intersection(int *a,int a_len,int *b,int b_len,int *result){int i = 0;int j = 0;int k = 0;                         //1,3,5,7,9,11int r_len;                         //1,2,3,4,5,6,7,8while(i < a_len && j < b_len){if(*(a + i) < *(b + j)){++i;}else if(*(a + i) > *(b + j)){++j;}else{//cout<<*(a + i)<<" ";*(result + k) = *(a + i);++k;++i;++j;}}if(k == 0)   {cout<<"此两个数组没有交集"<<endl;}else{for(i = 0; i < k; ++i){cout<<result[i]<<" ";}cout<<endl;}}int main(){int a[] = {1,3,5,7,9,11};int b[] = {1,2,3,4,5,6,7,8};int a_len = sizeof(a)/sizeof(int);int b_len = sizeof(b)/sizeof(int);int r_len = a_len >= b_len ? a_len : b_len;int *result = new int[r_len];Array_Intersection(a,a_len,b,b_len,result);return 0;}

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