LeetCode 1. Two Sum

Problem: 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.

Example:

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

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

该题最容易想到的暴力解法就是两层循环,逐对相加与target比较。其代价过大，代码为：

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

时间复杂度为O(n)的做法是只遍历一次vector，然后用hash表的find快速查找存储的数字。其代码如下：

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

已遇到多题巧用hash table(unordered_map)简化算法的方法，注意hash table的使用！