編程演算法 - 和為s的兩個數字 代碼(C)

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和為s的兩個數字 代碼(C)

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

題目: 輸入一個遞增排序的數組和一個數字s, 在數組中尋找兩個數, 使得它們的和正好是s.

如果有多對數位和等於s, 輸出任意一對即可.

排序數組, 則可以從兩端(即最大值, 最小值)開始進行尋找, 當和大於時, 則減少前端, 當和小於時, 則遞增尾端.

時間複雜度O(n).

代碼:

/* * 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);}

輸出:

num1 = 4 num2 = 11