1. Two Sum

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 the same element twice.Example:Given nums = [2, 7, 11, 15], target = 9,

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

我的解法

时间：14%
空间：60%

主要慢的原因：
for循环不适用于大型数据，比较慢，使用哈希表来查找可以显著提高程序运行速度

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

推荐解法

速度：97.81%
空间：12.60%

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> m;
        vector<int> res;
        for (int i = 0; i < nums.size(); ++i) {
            m[nums[i]] = i;
        }
        for (int i = 0; i < nums.size(); ++i) {
            int t = target - nums[i];
            if (m.count(t) && m[t] != i) {
                res.push_back(i);
                res.push_back(m[t]);
                break;
            }
        }
        return res;
    }
};