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中文文档

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]
Output: 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 <= 103
  • -109 <= nums[i] <= 109
  • -109 <= target <= 109
  • Only one valid answer exists.

Solutions

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, n = nums.length; i < n; ++i) {
            int num = target - nums[i];
            if (map.containsKey(num)) {
                return new int[]{map.get(num), i};
            }
            map.put(nums[i], i);
        }
        return null;
    }
}

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 [];
};

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