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1. 两数之和

English Version

题目描述

给定一个整数数组 nums 和一个整数目标值 target，请你在该数组中找出 和为目标值 的那 两个 整数，并返回它们的数组下标。

你可以假设每种输入只会对应一个答案。但是，数组中同一个元素在答案里不能重复出现。

你可以按任意顺序返回答案。

 

示例 1：

输入：nums = [2,7,11,15], target = 9
输出：[0,1]
解释：因为 nums[0] + nums[1] == 9 ，返回 [0, 1] 。

示例 2：

输入：nums = [3,2,4], target = 6
输出：[1,2]

示例 3：

输入：nums = [3,3], target = 6
输出：[0,1]

 

提示：

  • 2 <= nums.length <= 103
  • -109 <= nums[i] <= 109
  • -109 <= target <= 109
  • 只会存在一个有效答案

解法

用哈希表（字典）存放数组值以及对应的下标。

遍历数组，当发现 target - nums[i] 在哈希表中，说明找到了目标值。

Python3

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

Java

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

C++

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> map;
        for (int i = 0; i < nums.size(); ++i) {
            int num = target - nums[i];
            if (map.find(num) != map.end()) {
                return {map[num], i};
            }
            map[nums[i]] = i;
        }
        return {};
    }
};

Go

func twoSum(nums []int, target int) []int {
	numMap := make(map[int]int)
	for i, num := range nums {
		other := target - num
		if _, ok := numMap[other]; ok {
			return []int{numMap[other], i}
		}
		numMap[num] = i
	}
	return nil
}

JavaScript

var twoSum = function (nums, target) {
    const map = new Map();
    for (let i = 0; i < nums.length; i++) {
        if (map.has(target - nums[i])) {
            return [map.get(target - nums[i]), i];
        }
        map.set(nums[i], 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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