LeetCode 0167. Two Sum II - Input array is sorted两数之和 II - 输入有序数组【Easy】【Python】【双指针】
题目
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.
翻译
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 。
思路
双指针
left 指针从头指向尾，right 指针从尾指向头，然后判断两数之和是否等于 target。
时间复杂度：O(n)
Python代码
class Solution(object):
def twoSum(self, numbers, target):
"""
:type numbers: List[int]
:type target: int
:rtype: List[int]
"""
left = 0 # 从头指向尾
right = len(numbers) - 1 # 从尾指向头
while left < right:
if numbers[left] + numbers[right] == target:
return [left + 1, right + 1]
elif numbers[left] + numbers[right] > target:
right -= 1
else:
left += 1
return []
