README_EN.md

1. Two Sum

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Description

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.

 

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

 

Constraints:

• 2 <= nums.length <= 104
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• Only one valid answer exists.

 

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

Solutions

Python3

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        m = {}
        for i, v in enumerate(nums):
            x = target - v
            if x in m:
                return [m[x], i]
            m[v] = i

Java

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> m = new HashMap<>();
        for (int i = 0; i < nums.length; ++i) {
            int v = nums[i];
            int x = target - v;
            if (m.containsKey(x)) {
                return new int[] {m.get(x), i};
            }
            m.put(v, i);
        }
        return null;
    }
}

C++

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> m;
        for (int i = 0; i < nums.size(); ++i) {
            int v = nums[i];
            int x = target - v;
            if (m.count(x)) return {m[x], i};
            m[v] = i;
        }
        return {};
    }
};

Go

func twoSum(nums []int, target int) []int {
	m := map[int]int{}
	for i, v := range nums {
		x := target - v
		if j, ok := m[x]; ok {
			return []int{j, i}
		}
		m[v] = i
	}
	return nil
}

C#

public class Solution {
    public int[] TwoSum(int[] nums, int target) {
        var m = new Dictionary<int, int>();
        for (var i = 0; i < nums.Length; ++i)
        {
            int j;
            int v = nums[i];
            int x = target - v;
            if (m.TryGetValue(x, out j))
            {
                return new [] {j, i};
            }
            if (!m.ContainsKey(v))
            {
                m.Add(v, i);
            }

        }
        return null;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (nums, target) {
    const m = new Map();
    for (let i = 0; i < nums.length; ++i) {
        const v = nums[i];
        const x = target - v;
        if (m.has(x)) {
            return [m.get(x), i];
        }
        m.set(v, i);
    }
    return [];
};

Swift

class Solution {
    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var map = [Int: Int]()
        var i = 0
        for num in nums {
            map[num] = i
            i = i + 1
        }
        i = 0
        for num in nums {
            if let otherIndex = map[target - num], otherIndex != i {
                return [i, otherIndex]
            }
            i = i + 1
        }
        return []
    }
}

Nim

import std/enumerate

proc twoSum(nums: seq[int], target: int): seq[int] =
    var
        bal: int
        tdx: int
    for idx, val in enumerate(nums):
        bal = target - val
        if bal in nums:
            tdx = nums.find(bal)
            if idx != tdx:
                return @[idx, tdx]

Rust

use std::collections::HashMap;

pub fn soluation(nums: Vec<i32>, target: i32) -> Vec<i32> {
    let mut map = HashMap::new();
    for (i, item) in nums.iter().enumerate() {
        if map.contains_key(item) {
            return vec![i as i32, map[item]];
        } else {
            let x = target - nums[i];
            map.insert(x, i as i32);
        }
    }
    unreachable!()
}

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