1 问题
输入一个递增排序数组和数字和s,在数组里面找2个数，他们的和是s,如果有多对，只需要输出一对。
比如数组{1, 2, 4, 7, 11, 15}，我们输出4 ，11
2 思路
我们定义2个首尾指针，先是1+15，大于15，然后我们尾巴指针左移一下，然后就是1+11 小于15，然后首指针右移动一下，2+11，依次类推。
3 代码实现
#include <stdio.h>
#include <stdlib.h>#define true 1
#define false 0int findNumber(int *a, int len, int sum, int *num1, int *num2)
{int result = false; if (NULL == a || len < 1 || NULL == num1 || NULL == num2){return result;}int start = 0;int end = len - 1;while (end > start){*num1 = a[start];*num2 = a[end];int curSum = *num1 + *num2;if (curSum == sum){result = true;break;}else if (curSum > sum){--end;}else{++start;}}return result;
}int main()
{int a[] = {1, 2, 4, 7, 11, 15};int num1 = 1;int num2 = 3;int result = findNumber(a, sizeof(a) / sizeof(int), 15, &num1, &num2);if (result == 1){printf("the num1 is %d the num2 is %d\n", num1, num2);}else{printf("do not find the two number in numbers");}return 0;
}
4 运行结果
the num1 is 4 the num2 is 11
5 代码优化
while (end > start){int curSum = a[start] + a[end];if (curSum == sum){*num1 = a[start];*num2 = a[end];result = true;break;}else if (curSum > sum){--end;}else{++start;}}
