Leetcode: 1.Two Sum

Description

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].

思路

• 直接一个排序，O(nlgn)，然后一个查找，O(n)
• 使用unordered_map<int,int>可直接实现O(n)的解法

代码

• O(nlgn)的解法

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res;
        int len = nums.size();
        if(len == 0) return res;
        
        vector<pair<int, int>> vec;
        for(int i = 0; i < len; ++i){
            vec.push_back(pair<int, int>(nums[i], i));
        }
        
        sort(vec.begin(), vec.end(), 
            [](const pair<int, int> &a, const pair<int, int>&b){ return a.first < b.first;});
            
        int i = 0, j = len - 1, tmp;
        while(i < j){
            tmp = vec[i].first + vec[j].first;
            if(tmp == target){
                res.push_back(min(vec[i].second, vec[j].second));
                res.push_back(max(vec[i].second, vec[j].second));
                break;
            }
            else if(tmp < target)
                i++;
            else j--;
        }
        
        return res;
    }
};

• O(n)的解法

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res;
        int len = nums.size();
        if(len == 0) return res;
        
        unordered_map<int, int> hash;
        int needToFind = 0;
        for(int i = 0; i < len; ++i){
            needToFind = target - nums[i];
            if(hash.find(needToFind) != hash.end()){
                res.push_back(min(i, hash[needToFind]));
                res.push_back(max(i, hash[needToFind]));
                break;
            }
            else{
                hash.insert(pair<int, int>(nums[i], i));
            }
        }
        return res;
    }
};