Two Sum II

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution and you may not use the same element twice.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

翻译：给出一个升序排列的整数数组，找到两个数，加起来为目标数字

该函数返回加起来为目标数字的两个数的下标，其中index1小于index2.且index1和index2都不是从0开始（从1开始，所以数组的下标要加一再返回）

假设每个输入有且只有一个答案，相同元素不能使用两次。

输入：numbers={2, 7, 11, 15}, target=9

输出： index1=1, index2=2

大神答案（titanduan3）：

public int[] twoSum(int[] num, int target) {
    int[] indice = new int[2];
    if (num == null || num.length < 2) return indice;
    int left = 0, right = num.length - 1;
    while (left < right) {
        int v = num[left] + num[right];
        if (v == target) {
            indice[0] = left + 1;
            indice[1] = right + 1;
            break;
        } else if (v > target) {
            right --;
        } else {
            left ++;
        }
    }
    return indice;
}

我自己是用hashmap做的，膜拜大神的答案，思想很重要啊~~~