Leetcode 1.两数之和

一、Leetcode之两数之和

三种解法，其他实现可以参考leetcode解答

1.暴力法 2.二分法 3. 哈希

给定num，在对应数组中找到对应的两个数。

class Solution:    
    def twoSum1(self, nums: List[int], target: int) -> List[int]:
        length = len(nums)
        for i in range(length):
            for j in range(i+1, length):
                if nums[i] + nums[j] == target:
                    return [i, j]


    def twoSum2(self, nums: List[int], target: int) -> List[int]:
        length = len(nums)
        num_2_idx = {}     # num_2_idx = {num:idx for idx, num in enumerate(nums)} # dict会覆盖
        for i in range(length):
            if not num_2_idx.__contains__(nums[i]):
                num_2_idx[nums[i]] = i
        ori_nums = nums
        nums = sorted(nums)  # 原有的索引已经改变了
        for i in range(length):
            left = i + 1
            right = length - 1
            complement = target - nums[i]
            while left <= right:
                j = (left + right) // 2  # mid  
                if nums[j] > complement:
                    right = j - 1
                elif nums[j] < complement:  
                    left = j + 1      
                elif nums[j] == complement:
                    i_idx = num_2_idx[nums[i]]  
                    j_idx = num_2_idx[nums[j]]             
                    if nums[i] == nums[j]:                    
                        j_idx = i_idx + 1 + ori_nums[i_idx+1:].index(nums[j])
                    return [i_idx, j_idx] # 返回的是排序前元素的索引 


    def twoSum3(self, nums: List[int], target: int) -> List[int]:  
        num_2_idx = {} # 相当于逆序的暴力法
        for i in range(len(nums)):
            complement = target - nums[i]
            if num_2_idx.__contains__(complement): 
                return [num_2_idx.get(complement), i]                     
            num_2_idx[nums[i]] = i


    def twoSum(self, nums: List[int], target: int) -> List[int]:
        return self.twoSum2(nums, target)