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two-sum

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Two Sum – LeetCode 1

Problem

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].

Answer

Original

Code

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

思路

从数组nums的第一项开始寻找能加和成为target的另一项，简单粗暴。时间复杂度$O(n^2$)，空间复杂度$O(1)$。
耗时$189$ ms，排名$77.08\%$。

Better

思路

利用HashMap中查找操作的理论时间复杂度仅为$O(1)$来省去嵌套遍历的$O(n)$。通过建立一张表来对应所需要的nums[i] - target这一值，是以空间换时间的手段。 时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$9$ ms，排名$45.75\%$。

Code

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class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int> val2index;
            
        // Map first index to value
        val2index[nums[0]] = 0;
        
        for (int i=1; i < nums.size(); ++i) {
            // Find Complements in map
            int complement = target - nums[i];
            unordered_map<int,int>::const_iterator found = val2index.find(complement);
            if (found != val2index.end()) {
                return vector<int>({found->second,i});
            }
            
            // Map all index of nums to values
            val2index[nums[i]] = i;
        }
    }
};