【LeetCode】167. Two Sum II - Input array is sorted-优快云博客

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167. Two Sum II - Input array is sorted

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.

Note:

Your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.

Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

解题思路

（1）双指针解法，时间复杂度为 $O( n)$

已经AC的代码：

from typing import List


class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        if len(numbers) <= 0:
            return None
        low = 0
        high = len(numbers) - 1
        while low < high:
            if numbers[low] + numbers[high] == target:
                return [low + 1, high + 1]
            elif numbers[low] + numbers[high] > target:
                high -= 1
            else:
                low += 1


if __name__ == "__main__":
    numbers = [2, 7, 11, 15]
    target = 9
    sol = Solution()
    print(sol.twoSum(numbers, target))

（2）暴力解法（双重循环）,时间复杂度为 $O(n^2)$

通过两次循环，每次循环遍历列表，这种算法的时间复杂度是 $O(n^2)$ 。

代码提交后：Time Limit Exceeded，代码如下：

    # 方法二：暴力解法（双重循环）
    def twoSum2(self, numbers: List[int], target: int) -> List[int]:
        if len(numbers) <= 0:
            return None

        for i in range(len(numbers)):
            for j in range(i + 1, len(numbers)):
                if numbers[i] + numbers[j] == target:
                    return [i + 1, j + 1]

（3）二分搜索法，时间复杂度为 $O(n log n)$

数组是有序的，那么我们很容易想到使用二分搜索法是不是可以用于这个问题呢？我们可以每次判断 $target-num[i]$ 对应的值是否在 $num[i+1:]$ 中，这个时候算法的复杂度变成了 $O(nlogn)$ 。

已经AC的代码如下：

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        length = len(numbers)
        for i in range(length):
            difference = target - numbers[i]
            index = self.search_Bin(numbers, difference, i + 1)
            if index != -1:
                return [i + 1, index + 1]

    def search_Bin(self, numbers, target, start):
        if len(numbers) <= 0:
            return -1
        low = start
        high = len(numbers) - 1
        while low <= high:
            middle = (low + high) // 2
            if numbers[middle] == target:
                return middle
            elif numbers[middle] > target:
                high = middle - 1
            else:
                low = middle + 1
        return -1