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TwoSumIIInputArrayIsSorted.py
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70 lines (57 loc) · 2.19 KB
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#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time : 2019/11/5 19:25
# @Author : tc
# @File : TwoSumII InputArrayIsSorted.py
"""
题号167 两数之和 II - 输入有序数组
给定一个已按照升序排列 的有序数组,找到两个数使得它们相加之和等于目标数。
函数应该返回这两个下标值 index1 和 index2,其中 index1 必须小于 index2。
说明:
返回的下标值(index1 和 index2)不是从零开始的。
你可以假设每个输入只对应唯一的答案,而且你不可以重复使用相同的元素。
示例:
输入: numbers = [2, 7, 11, 15], target = 9
输出: [1,2]
解释: 2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。
本题可以用双指针和二分查找两种方法
参考:https://leetcode-cn.com/problems/two-sum-ii-input-array-is-sorted/solution/shuang-zhi-zhen-dui-zhuang-er-fen-fa-python-dai-ma/
"""
from typing import List
class Solution:
# 双指针
def twoSum(self, numbers: List[int], target: int) -> List[int]:
if not numbers:
return []
left = 0
right = len(numbers) - 1
while left < right:
if numbers[left] + numbers[right] == target:
return [left+1, right+1]
elif numbers[left] + numbers[right] < target:
left += 1
else:
right -= 1
return []
# 二分查找
def twoSum2(self, numbers: List[int], target: int) -> List[int]:
size = len(numbers)
for left in range(size - 1):
right = self.binary_search(left+1,size - 1,numbers,target-numbers[left])
if right != -1:
return [left+1,right+1]
def binary_search(self,left,right,numbers,target):
while left < right:
mid = left + ((right - left)>>1)
if numbers[mid] == target:
right = mid
elif numbers[mid] > target:
right = mid
elif numbers[mid] < target:
left = mid + 1
return left if numbers[left] == target else -1
if __name__ == '__main__':
numbers = [2, 7, 11, 15]
target = 9
solution = Solution()
print(solution.twoSum2(numbers,target))