【LeetCode】两数之和 II - 输入有序数组

韩旭051

发布于 2019-12-03 01:20:18

2060

发布于 2019-12-03 01:20:18

文章被收录于专栏：刷题笔记刷题笔记

本文链接：https://blog.csdn.net/shiliang97/article/details/103188270

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.

有个大数会卡一下，先看看我写的（辣鸡死了）

while(numbers[i]==numbers[i+1]&&numbers[i]*2!=target)i++;

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
         vector<int> v;
        for(int i=0;i<numbers.size();i++){
            while(numbers[i]==numbers[i+1]&&numbers[i]*2!=target)i++;
            int t=target-numbers[i];
            for(int l=i+1;l<numbers.size();l++){
                if(numbers[l]==t){
                    v.push_back(i+1);
                    v.push_back(l+1);
                    return v;
                }
            }
        }return v;
    }
};

人家其他人，两头堵~~~~就不是n2超时了

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

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原始发表：2019-11-30 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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