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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 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。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 <= 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

我们可以用哈希表 $m$ 存放数组值以及对应的下标。

遍历数组 nums，当发现 target - nums[i] 在哈希表中，说明找到了目标值，返回 target - nums[i] 的下标以及 $i$ 即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 nums 的长度。

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        m = {}
        for i, x in enumerate(nums):
            y = target - x
            if y in m:
                return [m[y], i]
            m[x] = i
class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> m = new HashMap<>();
        for (int i = 0;; ++i) {
            int x = nums[i];
            int y = target - x;
            if (m.containsKey(y)) {
                return new int[] {m.get(y), i};
            }
            m.put(x, i);
        }
    }
}
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> m;
        for (int i = 0;; ++i) {
            int x = nums[i];
            int y = target - x;
            if (m.count(y)) {
                return {m[y], i};
            }
            m[x] = i;
        }
    }
};
func twoSum(nums []int, target int) []int {
	m := map[int]int{}
	for i := 0; ; i++ {
		x := nums[i]
		y := target - x
		if j, ok := m[y]; ok {
			return []int{j, i}
		}
		m[x] = i
	}
}
function twoSum(nums: number[], target: number): number[] {
    const m: Map<number, number> = new Map();

    for (let i = 0; ; ++i) {
        const x = nums[i];
        const y = target - x;

        if (m.has(y)) {
            return [m.get(y)!, i];
        }

        m.set(x, i);
    }
}
use std::collections::HashMap;

impl Solution {
    pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let mut m = HashMap::new();
        for (i, &x) in nums.iter().enumerate() {
            let y = target - x;
            if let Some(&j) = m.get(&y) {
                return vec![j as i32, i as i32];
            }
            m.insert(x, i as i32);
        }
        unreachable!()
    }
}
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (nums, target) {
    const m = new Map();
    for (let i = 0; ; ++i) {
        const x = nums[i];
        const y = target - x;
        if (m.has(y)) {
            return [m.get(y), i];
        }
        m.set(x, i);
    }
};
public class Solution {
    public int[] TwoSum(int[] nums, int target) {
        var m = new Dictionary<int, int>();
        for (int i = 0, j; ; ++i) {
            int x = nums[i];
            int y = target - x;
            if (m.TryGetValue(y, out j)) {
                return new [] {j, i};
            }
            if (!m.ContainsKey(x)) {
                m.Add(x, i);
            }
        }
    }
}
class Solution {
    /**
     * @param Integer[] $nums
     * @param Integer $target
     * @return Integer[]
     */
    function twoSum($nums, $target) {
        $m = [];
        foreach ($nums as $i => $x) {
            $y = $target - $x;
            if (isset($m[$y])) {
                return [$m[$y], $i];
            }
            $m[$x] = $i;
        }
    }
}
import scala.collection.mutable

object Solution {
  def twoSum(nums: Array[Int], target: Int): Array[Int] = {
    var map = new mutable.HashMap[Int, Int]()
    for (i <- 0 to nums.length) {
      if (map.contains(target - nums(i))) {
        return Array(map(target - nums(i)), i)
      } else {
        map += (nums(i) -> i)
      }
    }
    Array(0, 0)
  }
}
class Solution {
    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var m = [Int: Int]()
        var i = 0
        while true {
            let x = nums[i]
            let y = target - nums[i]
            if let j = m[target - nums[i]] {
                return [j, i]
            }
            m[nums[i]] = i
            i += 1
        }
    }
}
# @param {Integer[]} nums
# @param {Integer} target
# @return {Integer[]}
def two_sum(nums, target)
  nums.each_with_index do |x, idx|
    if nums.include? target - x
      return [idx, nums.index(target - x)] if nums.index(target - x) != idx
    end
    next
  end
end
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]