寻找满足条件的两个数

  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