LeetCode 167. Two Sum II - Input array is sorted--Python解法

题目地址：Two Sum II - Input array is sorted - LeetCode

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.

这道题目最容易想到的是穷举法，但肯定会超时，而且没有利用有序这个特性。
然后就可以想到穷举的时候进行二分查找，这样就利用了有序这个特性。

Python解法如下：

class Solution:def twoSum(self, numbers: list, target: int) -> list:result = [0, 0]length = len(numbers)flag = 0def search(begin, end, number):nonlocal numbers, result, flagif begin > end:returnmiddle = (begin+end)//2if numbers[middle] == number:flag = 1result[1] = middle+1returnelif numbers[middle] < number:search(middle+1, end, number)else:search(begin, middle-1, number)for i in range(0, length-1):search(i+1, length-1, target-numbers[i])if flag == 1:result[0] = i+1return result

然后把递归的二分查找改为迭代的二分查找：

class Solution:def twoSum(self, numbers: list, target: int) -> list:result = [0, 0]length = len(numbers)flag = 0def search(begin, end, number):nonlocal numbers, result, flagwhile begin <= end:middle = (begin+end)//2if numbers[middle] == number:flag = 1result[1] = middle+1returnelif numbers[middle] < number:begin = middle+1else:end = middle-1for i in range(0, length-1):search(i+1, length-1, target-numbers[i])if flag == 1:result[0] = i+1return result

可以通过所有样例了，但结果不让人满意，随后可以想到既然是有序的，可以从两边向中间逼近：

class Solution:def twoSum(self, numbers: list, target: int) -> list:result = [0, 0]length = len(numbers)begin = 0end = length-1while begin < end:temp = numbers[begin]+numbers[end]if temp == target:result[0] = begin+1result[1] = end+1return resultelif temp > target:end -= 1else:begin += 1

这样时间复杂度就变为了O(N)。

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