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Description

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.

一.题目理解

这道题和第1题极像，区别在于给定的数组已经排好序，输出要求下标加1且顺序递增。

二.题目解答

这里直接给出代码

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {

        vector<int> ans;
        unordered_map<int,int> umap;

        for(int i = 0; i < numbers.size(); i++) {

            if(umap.count(target - numbers[i])) {

                ans.push_back(umap[target - numbers[i]] + 1);
                ans.push_back(i+1);
                
                return ans;

            }
            umap[numbers[i]] = i;

        }

        return ans;

    }
};

167.Two Sum II–Input is sorted

标签：i++   border   turned   esc   递增   xpl   count   dex   col   

原文地址：https://www.cnblogs.com/ovs98/p/9846014.html