Leetcode - 1 - Two Sum

1. Two Sum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Solution:

One-pass Hash Table

 

Python

class Solution(object):
    def twoSum(self, nums, target):
        if len(nums) <= 1:
            return False
        buff_dict = {}
        for i in range(len(nums)):
            if nums[i] in buff_dict:
                return [buff_dict[nums[i]], i]
            else:
                buff_dict[target - nums[i]] = i

https://leetcode.com/problems/two-sum/discuss/17/Here-is-a-Python-solution-in-O(n)-time

 

Java

public int[] twoSum(int[] numbers, int target) {
    int[] result = new int[2];
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    for (int i = 0; i < numbers.length; i++) {
        if (map.containsKey(target - numbers[i])) {
            result[1] = i;
            result[0] = map.get(target - numbers[i]);
            return result;
        }
        map.put(numbers[i], i);
    }
    return result;
}

https://leetcode.com/problems/two-sum/discuss/3/Accepted-Java-O(n)-Solution

 

C++

vector<int> twoSum(vector<int> &numbers, int target)
{
    //Key is the number and value is its index in the vector.
	unordered_map<int, int> hash;
	vector<int> result;
	for (int i = 0; i < numbers.size(); i++) {
		int numberToFind = target - numbers[i];

            //if numberToFind is found in map, return them
		if (hash.find(numberToFind) != hash.end()) {
			result.push_back(hash[numberToFind]);
			result.push_back(i);			
			return result;
		}

            //number was not found. Put it in the map.
		hash[numbers[i]] = i;
	}
	return result;
}

https://leetcode.com/problems/two-sum/discuss/13/Accepted-C%2B%2B-O(n)-Solution

Changed in result.push_back, with '+1' deleted.

In c++, finding in hash needs to be written as hash.find(tofind) != hash.end()