#1 Two Sum[E]

Description

tags: Array, Hash Table
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].

My solution

class Solution {
public:
    static vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> result;
        map<int, vector<int>> numMap;
        // store nums in a map
        for (auto i = 0; i < nums.size(); i++)
        {
            numMap[nums[i]].push_back(i);
        }

        for (map<int, vector<int>>::iterator it = numMap.begin(); it != numMap.end(); it++)
        {
            int firstVal = it->first;
            int secondVal = target - firstVal;
            if (firstVal == secondVal)
            {
                if (it->second.size() > 1)
                {
                    result.push_back(it->second[0]);
                    result.push_back(it->second[1]);
                    return result;
                }
                continue;
            }
            else
            {
                if (numMap.find(secondVal) != numMap.end())
                {
                    result.push_back(it->second[0]);
                    result.push_back(numMap[secondVal][0]);
                    return result;
                }
                continue;
            }
        }
        return result;
    }
};

Analysis

  • Time Complexity: O(nlogn)
    Space Complexity: O(n)
  • cons:
    • Not concise enough
    • Time Complexity can cut down to O(n) using hash table

Points

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