【c/c++】 leetcode 167. Two Sum II - Input array is sorted （easy）

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/maotianyi941005/article/details/84948846

 167. Two Sum II - Input array is sorted （easy）

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.

Note:

• 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.

Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

思路

一个start和一个end位置指向首尾，如果（numbers[start] + numbers[end] ）> target,说明要减小左边的值，所以减小end的位置；

如果（numbers[start] + numbers[end] ）< target,说明要增大左边的值，所以增大start的位置；

这道题简单在只有一对满足所求条件 且 vector是增序

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

4ms beats 99.28%