167. Two Sum II - Input array is sorted

问题描述

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

题目链接：

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思路分析

给一个升序数组，其中有且只有一对元素的和未target，找到这两个元素下标。

使用两个指针，分别从数组的头尾进行遍历。当两指针元素的和小于target时，low++（反过来也一样），这样可以保证有一个指针先到达目标下标处，然后在操作另一个指针，利用了只有一对元素的和为target的题目条件。

代码

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int sum;
        int low = 0, high =  numbers.size() - 1;
        vector<int> result;
        while(low < high){
            sum = numbers[low] + numbers[high];
            if (sum == target){
                result.push_back(low + 1);
                result.push_back(high + 1);
                return result;
            }
            if (sum < target)
                low++;
            else
                high--;
        }
    }
};

时间复杂度：O(n);

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反思

这个题目蛮么意思的，暴力破解因为测试用例的原因TL了，必须利用必定有一对元素的和为target的特性来入手。