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 
#include 
#include 

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