寻找满足条件的两个数

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一：输入一个数组和一个数字，在数组中查找两个数，使得它们的和正好是输入的那个数字

代码如下：

#include <iostream>#include <cstdlib>using namespace std;struct m_elem{int unm;int data;};struct m_pair{int f, s;m_pair* next;m_pair(int ff, int ss) : f(ff), s(ss){ next = NULL;}};int cmp(const void* a, const void* b){m_elem* f = (m_elem*)a;m_elem* s = (m_elem*)b;return f->data - s->data;}m_pair* m_findSum(int arr[], const int nLen, int sum){m_elem m_arr[nLen];for(int i = 0; i < nLen; i++){m_arr[i].unm = i;m_arr[i].data = arr[i];}qsort(m_arr, nLen, sizeof(m_elem), cmp);int left = 0, right = nLen-1;m_pair* retP = NULL;while(left < right){if(m_arr[left].data + m_arr[right].data < sum){++left;}else if(m_arr[left].data + m_arr[right].data > sum){--right;}else{m_pair* t = new m_pair(m_arr[left].unm, m_arr[right].unm);if(retP == NULL){retP = t;}else{t->next = retP;retP = t;}++left; ++right;continue;}}return retP;}int main(int argc, char* argv[]){int sum = 50;int testArr[] = {3, 46, 43, 4, 2, 5, 47, 1, 45, 109};const int len = sizeof(testArr)/sizeof(int);m_pair* r = m_findSum(testArr, len, sum);if(r != NULL){int i = 0;m_pair* tmp;while(r != NULL){tmp = r;cout << "第" << ++i << "组: " << r->f << " " << r->s << endl;r = r->next;delete tmp;}}else{cout << "数组中没有两个数的和为：" << sum << endl;}return 0;}

附：输出多对符合条件的数对在原始数组中的位置

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