파이썬 대 Golang 대 러스트

테스트 시나리오

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.

구현

파이썬

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        hash_table = {}
        for i, num in enumerate(nums):
            target_num = target - num
            if num in hash_table:
                return i, hash_table[num]
            else:
                hash_table[target_num] = i
        return None

파이썬 통계

골랑

func twoSum(nums []int, target int) []int {
    hashMap := make(map[int] int)
    for i := 0; i < len(nums); i++{
        if _, found := hashMap[nums[i]]; found {
            ans := []int{i, hashMap[nums[i]]}
            return ans
        } else {
            hashMap[target- nums[i]]= i
        }
    }
    return nil
}

골랑 통계

녹

use std::collections::HashMap;
impl Solution {
    pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let mut hash_table: HashMap<i32, i32> = HashMap::new();
    for i in 0..nums.len() {
        // println!("Processing number: {}", nums[i]);
        match hash_table.get(&nums[i]){
            Some(&x) => return vec![x, i as i32],
            None => hash_table.insert(target - nums[i], i as i32),
        };
    };
    return vec![-1, -1]
    }
}

녹 통계

결론