Two Sum最优解法

链接: https://leetcode.com/problems/two-sum
思路:
建立一个哈希表存放所有数字，迭代访问所有数字， 每次都尝试取出另一个数字的位置，如果成功返回这两个数字位置, 否则将数字的位置存到哈希表。这样能完美解决两个数字相等的情况。时间复杂度和空间复杂度都为o(n). 解决方案基于链接: https://leetcode.com/problems/two-sum/discuss/1610384/Optimal-Solution-Explained
解法:

class Solution:

    def twoSum(self, nums: List[int], target: int) -> List[int]:
        prevMap = {}
        for i, n in enumerate(nums):
            diff = target - n
            if diff in prevMap:
                return [prevMap[diff], i]
            prevMap[n] = i
        return []

另外还有一种方法是排序，然后遍历排序后的数字，对另一个数字进行二分搜索。这种方式时间复杂度为o(nlogn), 空间复杂度为o(n)。

class Solution:

    def binarySearch(self, nums, lower, uppuer, target):
        if lower > uppuer:
            return -1
        mid = lower + int((uppuer - lower) / 2)
        if nums[mid] > target:
            return self.binarySearch(nums, lower, mid - 1, target)
        if nums[mid] < target:
            return self.binarySearch(nums, mid + 1, uppuer, target)
        return mid

    def twoSum(self, nums: List[int], target: int) -> List[int]:
        sorted_numbers = sorted(nums)
        len_nums = len(nums)
        len_nums_minus_1 = len_nums - 1
        for i in range(len_nums):
            left = sorted_numbers[i]
            right = target - left
            index = self.binarySearch(sorted_numbers, i + 1, len_nums_minus_1, right)
            if index != -1:
                left_index = 0
                right_index = 0
                j = 0
                if left != right:
                    while j < len_nums:
                        if nums[j] == left:
                            left_index = j
                        if nums[j] == right:
                            right_index = j
                        j += 1
                    return [left_index, right_index]
                while j < len_nums:
                    if nums[j] == left:
                        left_index = j
                    j += 1
                j = len_nums_minus_1
                while j >= 0:
                    if nums[j] == right:
                        right_index = j
                    j -= 1
                return [left_index, right_index]
                    
        return []