leetcode-1 - Two Sum

動機

lambda加map好方便

Problem

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 = 9Output: [0,1]Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6Output: [0,1]

Constraints:

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

Sol

先把原本的index加到list，之後sort 做two pointer

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        nums = [(n,i) for (i,n) in enumerate(nums)]
        nums.sort(key=lambda a: a[0])
        i,j = [0, len(nums)-1]
        while i < j:
            if nums[i][0]+nums[j][0] == target:
                return [nums[i][1], nums[j][1]]
            elif nums[i][0]+nums[j][0] >= target:
                j -= 1
            else:
                i += 1