LeetCode 1.Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

方法1：直接暴力求解，时间复杂度为O（n^2）,可以AC，代码如下：

vector<int> twoSum(vector<int>& nums, int target) {
        int len = nums.size();
        vector<int>ans;
        此方法为暴力，时间复杂度为O(n^2)
        for (int i = 0; i < len; i++){
            int tmp = target - nums[i];
            for (int j = i+1; j < len; j++){
                if (tmp == nums[j]){
                    ans.push_back(i);
                    ans.push_back(j);
                }
            }
        }}
        return ans;
    }

方法2：利用Hash表求解，先保存每一个元素的下标，时间复杂度为O（n），代码如下：

vector<int> twoSum(vector<int>& nums, int target) {
        int len = nums.size();
        unordered_map<int,int>mapping;
        vector<int>ans;
        for (int i = 0; i < len; i++){
            mapping[nums[i]] = i;//记录每个元素的下标
        }
        for (int i = 0; i < len; i++){
            int tmp = target - nums[i];
            if (mapping.find(tmp) != mapping.end() && mapping[tmp] > i)
            {
                ans.push_back(i);
                ans.push_back(mapping[tmp]);
                break;
            }
        }
        return ans;
    }