【LeetCode】1. Two Sum

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].

方法1：暴力解

• Time complexity : $O\left ( n^{2} \right )$O(n2)
• Space complexity : $O\left ( 1 \right )$O(1)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        for(int i = 0 ; i < nums.size() ; i ++)
            for(int j = i + 1 ; j < nums.size() ; j ++)
                if(nums[i] + nums[j] == target){
                    int res[] = {i, j};
                    return vector<int>(res, res + 2);
                }
    }
};

方法2：两次遍历

• Time complexity : $O\left ( n \right )$O(n)
• Space complexity : $O\left ( n \right )$O(n)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int> hash;
        for(int i = 0 ; i < nums.size() ; i ++)
            hash[nums[i]] = i;

        for(int i = 0 ; i < nums.size() ; i ++){
            int complement = target - nums[i];
            if(hash.find(complement) != hash.end() && hash[complement] != i){
                int res[] = {i, hash[complement]};
                return vector<int>(res, res + 2);
            }
        }
    }
};

方法3：一次遍历

• Time complexity : $O\left ( n \right )$O(n)
• Space complexity : $O\left ( n \right )$O(n)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int> hash;
        for(int i = 0 ; i < nums.size() ; i ++){
            int complement = target - nums[i];
            if(hash.find(complement) != hash.end()){
                int res[] = {i, hash[complement]};
                return vector<int>(res, res + 2);
            }
            hash[nums[i]] = i;
        }
    }
};