LeetCode算法系列: 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].

 

solution 1:

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> res;
        for(int i = 0; i < nums.size(); i++)
        {
            for(int j = i+1; j < nums.size(); j++)
            {
                if((target - nums[i]) ==nums[j])
                    res.push_back(i), res.push_back(j);                
            }
        }
        if(!res.empty()) return res;
        else printf("no solution");
    }
};

空间复杂度为O(1)，时间复杂度为O(n^2)

solution 2:

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 another = target - nums[i];
            
            if(m.count(another) && m[another] != i)
            {
                res.push_back(i);
                res.push_back(m[another]);
                break;
            }
        }
        if(!res.empty()) return res;
        else printf("no solution");
    }
};

空间复杂度为O(n)，时间复杂度为O(n)