编程算法 - 和为s的两个数字 代码(C)

和为s的两个数字 代码(C)

本文地址: http://blog.csdn.net/caroline_wendy

题目: 输入一个递增排序的数组和一个数字s, 在数组中查找两个数, 使得它们的和正好是s.
如果有多对数字的和等于s, 输出任意一对即可.

排序数组, 则可以从两端(即最大值, 最小值)开始进行查找, 当和大于时, 则减少前端, 当和小于时, 则递增尾端.
时间复杂度O(n).

代码:
/*
* main.cpp
*
* Created on: 2014.6.12
* Author: Spike
*/

/*eclipse cdt, gcc 4.8.1*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

bool FindNumbersWithSum(int data[], int length, int sum, int* num1, int* num2)
{
bool found = false;
if (length<1 || num1==NULL || num2 == NULL)
return found;
int ahead = length-1;
int behind = 0;

while (ahead > behind) {
long curSum = data[ahead] + data[behind];
if (curSum == sum) {
*num1 = data[behind];
*num2 = data[ahead];
found = true;
break;
} else if (curSum < sum)
++behind;
else
--ahead;
}
return found;
}

int main(void)
{
int data[] = {1, 2, 4, 7, 11, 15};
int num1, num2;
if (!FindNumbersWithSum(data, 6, 15, &num1, &num2))
printf("Error\n");
printf("num1 = %d num2 = %d\n", num1, num2);
}

输出:
num1 = 4 num2 = 11