comments	difficulty	edit_url	tags
true	简单	https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0001.Two%20Sum/README.md	数组 哈希表

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 target  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，并且你不能使用两次相同的元素。

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 <= 104
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• 只会存在一个有效答案

 

进阶：你可以想出一个时间复杂度小于 O(n2) 的算法吗？

解法

方法一：哈希表

我们可以使用一个哈希表 $\mathit{\text{d}}$ 来存储每个元素及其对应的索引。

遍历数组 $\mathit{\text{nums}}$，对于当前元素 $\mathit{\text{nums}}[i]$，我们首先判断 $\mathit{\text{target}}-\mathit{\text{nums}}[i]$ 是否在哈希表 $\mathit{\text{d}}$ 中，如果在 $\mathit{\text{d}}$ 中，说明 $\mathit{\text{target}}$ 值已经找到，返回 $\mathit{\text{target}}-\mathit{\text{nums}}[i]$ 的索引和 $i$ 即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$，其中 $n$ 为数组 $\mathit{\text{nums}}$ 的长度。

Python3

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

Java

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

C++

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

Go

func twoSum(nums []int, target int) []int {
	d := map[int]int{}
	for i := 0; ; i++ {
		x := nums[i]
		y := target - x
		if j, ok := d[y]; ok {
			return []int{j, i}
		}
		d[x] = i
	}
}

TypeScript

function twoSum(nums: number[], target: number): number[] {
    const d = new Map<number, number>();
    for (let i = 0; ; ++i) {
        const x = nums[i];
        const y = target - x;
        if (d.has(y)) {
            return [d.get(y)!, i];
        }
        d.set(x, i);
    }
}

Rust

use std::collections::HashMap;

impl Solution {
    pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let mut d = HashMap::new();
        for (i, &x) in nums.iter().enumerate() {
            let y = target - x;
            if let Some(&j) = d.get(&y) {
                return vec![j as i32, i as i32];
            }
            d.insert(x, i);
        }
        vec![]
    }
}

JavaScript

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

C#

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

PHP

class Solution {
    /**
     * @param Integer[] $nums
     * @param Integer $target
     * @return Integer[]
     */
    function twoSum($nums, $target) {
        $d = [];
        foreach ($nums as $i => $x) {
            $y = $target - $x;
            if (isset($d[$y])) {
                return [$d[$y], $i];
            }
            $d[$x] = $i;
        }
    }
}

Scala

import scala.collection.mutable

object Solution {
    def twoSum(nums: Array[Int], target: Int): Array[Int] = {
        val d = mutable.Map[Int, Int]()
        var ans: Array[Int] = Array()
        for (i <- nums.indices if ans.isEmpty) {
            val x = nums(i)
            val y = target - x
            if (d.contains(y)) {
                ans = Array(d(y), i)
            } else {
                d(x) = i
            }
        }
        ans
    }
}

Swift

class Solution {
    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var d = [Int: Int]()
        for (i, x) in nums.enumerated() {
            let y = target - x
            if let j = d[y] {
                return [j, i]
            }
            d[x] = i
        }
        return []
    }
}

Ruby

# @param {Integer[]} nums
# @param {Integer} target
# @return {Integer[]}
def two_sum(nums, target)
    d = {}
    nums.each_with_index do |x, i|
      y = target - x
      if d.key?(y)
        return [d[y], i]
      end
      d[x] = i
    end
end

Kotlin

class Solution {
    fun twoSum(nums: IntArray, target: Int): IntArray {
        val m = mutableMapOf<Int, Int>()
        nums.forEachIndexed { i, x ->
            val y = target - x
            val j = m.get(y)
            if (j != null) {
                return intArrayOf(j, i)
            }
            m[x] = i
        }
        return intArrayOf()
    }
}

Nim

import std/enumerate
import std/tables

proc twoSum(nums: seq[int], target: int): seq[int] =
  var d = initTable[int, int]()
  for i, x in nums.pairs():
    let y = target - x
    if d.hasKey(y):
      return @[d[y], i]
    d[x] = i
  return @[]

Cangjie

class Solution {
    func twoSum(nums: Array<Int64>, target: Int64): Array<Int64> {
        let d = HashMap<Int64, Int64>()
        for (i in 0..nums.size) {
            if (d.contains(target - nums[i])) {
                return [d[target - nums[i]], i]
            }
            d[nums[i]] = i
        }
        []
    }
}