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.

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

使用第一題的方法依然是可行的，時間/空間復(fù)雜度是O(n)。

var twoSum = function(numbers, target) {
    var map = {};
    var num = numbers.length;
    for (var i = 0; i<num;i++) {
        if (map[target-numbers[i]]!==undefined)
            return [map[target-numbers[i]]+1,i+1];
        map[numbers[i]] = i;
    }
};

不過如果好好利用已經(jīng)被排為升序這個信息，我們可以找到更加優(yōu)化的算法。
因為數(shù)組已被排序，且一定存在解。那么我們可以從兩邊往中間找。

var twoSum = function(numbers, target) {
    var left = 0;
    var right = numbers.length-1;
    while(true) {
        if ((numbers[left]+numbers[right])===target) 
            return [left+1,right+1];
        else if ((numbers[left]+numbers[right])>target)
            right--;
        else
            left++;
    }
};