Given an array of integers, return indices of the two numbers such that they add up to a specific target.
给定一个integer类型的数组，让你返回数组中两个元素相加起来等于给定的值得元素的索引数组
You may assume that each input would have exactly one solution.
你可以假设每个给定值只有一个解决方案，那这里我们就不必考虑多个解决方案了，找到一个解决方案后直接返回结果。
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.
返回数组的索引已置为0
以下是我的解决方案：
int[] array = new int[2]; for (int i = 0; i < nums.Length - 1; i++) { int num1 = nums[i]; int j = i + 1; while(j < nums.Length) { int num2 = nums[j]; int result = num1 + num2; if (result.CompareTo(target) == 0) { array[0] = i; array[1] = j; i = j = nums.Length + 1; } j++; } } return array;
原文：http://www.cnblogs.com/chenyongblog/p/5225226.html