LeetCode第一题Two Sum的O(n)解法

leetcode刷题记录

第一题 Two Sum

题目描述：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.

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

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

解答：

初始想法是枚举法，对两个集合进行遍历，最终可以通过，但效率很低，只超过13%的人。这是万万不行的。

初始代码：

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

改进

使用HashTable implementation，可以将时间代价控制到O(n);

static int lambda_0 = []() { std::ios::sync_with_stdio(false); cin.tie(NULL); return 0; }();
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        
        vector<int> indices;
        map<int,int> valueIndices;
        

        
        for(int i =0; i < nums.size(); i++){
            
            if(valueIndices.find(target-nums[i]) != valueIndices.end()){
                
                    indices.push_back(i);
                    indices.push_back(valueIndices[target-nums[i]]);
                   
                    return indices;
            }
                    valueIndices[nums[i]] = i;
        }
         

    
    return indices;  
    }
};

时间复杂度减少到O(n),