【Leetcode刷题】算法:两数之和

一、题目描述

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二、尝试1

from typing import List
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        num_dict = {
    
    }

        for i in range(len(nums)):
            diff = target - nums[i]

            if diff in num_dict:
                return [num_dict[diff], i]

            num_dict[nums[i]] = i

        return []

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三、尝试2

from typing import List

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        num_dict = {
    
    }

        for i, num in enumerate(nums):
            diff = target - num
            
            if diff in num_dict:
                return [num_dict[diff], i]
            
            num_dict[num] = i
        
        return []

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四、尝试3

from typing import List

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        nums_sorted = sorted(enumerate(nums), key=lambda x: x[1])
        left, right = 0, len(nums) - 1

        while left < right:
            curr_sum = nums_sorted[left][1] + nums_sorted[right][1]

            if curr_sum == target:
                return [nums_sorted[left][0], nums_sorted[right][0]]

            if curr_sum < target:
                left += 1
            else:
                right -= 1

        return []

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五、尝试4

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        records = dict()

        # 用枚举更方便,就不需要通过索引再去取当前位置的值
        for idx, val in enumerate(nums):
            if target - val not in records:
                records[val] = idx
            else:
                return [records[target - val], idx]

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转载自blog.csdn.net/wzk4869/article/details/130728214