LeetCode-1-Two Sum

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Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly\ one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

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Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

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Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

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Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

  • 2 <= nums.length <= 104
  • -109 <= nums[i] <= 109
  • -109 <= target <= 109
  • Only one valid answer exists.

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

自己写的代码：

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class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> result;
        int e = nums.size()-1;
        int flag = 0;
        for (int i = 0; i <= e; i++) {
            for (int j = i+1; j <= e; j++) {
                if (nums[i] + nums[j] == target) {
                    result.push_back(i);
                    result.push_back(j);
                    flag = 1;
                }
            }
            if (flag)    break;
        }
        return result;
    }
};

别的大佬写的代码

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class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        map<int,int> a;//提供一对一的hash
        vector<int> b(2,-1);//用来承载结果，初始化一个大小为2，值为-1的容器b
        for(int i=0;i<nums.size();i++)
        {
            if(a.count(target-nums[i])>0)
            {
                b[0]=a[target-nums[i]];
                b[1]=i;
                break;
            }
            a[nums[i]]=i;//反过来放入map中，用来获取结果下标
        }
        return b;
    };
};

作者：陈乐乐
链接：https://leetcode.cn/problems/two-sum/solutions/4361/liang-shu-zhi-he-by-gpe3dbjds1/
来源：力扣（LeetCode）
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