NYUer | LeetCode1 Two Sum

LeetCode1 Two Sum


Author: Stefan Su
Create time: 2022-10-31 12:45:10
Location: New York City, NY, USA

Description Easy

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]
Constrains
  • 2 <= nums.length <= 104
  • -109 <= nums[i] <= 109
  • -109 <= target <= 109
  • Only one valid answer exists

Analysis

Use a for loop to go through all elements in the nums, and check current value and target - current value both in the unordered hash map record or not. If in it, return [current index, record[target - value of current index]]. If not, record[value of current index] = current index.

Solution

  • unordered hash map
class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        record = dict()

        # It's more convenient to use enumeration, you don't need to 
        # go through the index to get the value of the current position
        for index, value in enumerate(nums):
            if target - value not in record:
                record[value] = index
            else:
                return [index, record[target - value]]

Hopefully, this blog can inspire you when solving LeetCode1. For any questions, please comment below.

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转载自blog.csdn.net/Moses_SU/article/details/127626246