167.?Two Sum II - Input Array Is Sorted
Medium
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Given a?1-indexed?array of integers?numbers?that is already?sorted in non-decreasing order, find two numbers such that they add up to a specific?target?number. Let these two numbers be?numbers[index1]?and?numbers[index2]?where?1 <= index1?< index2?<= numbers.length.
Return?the indices of the two numbers,?index1?and?index2,?added by one?as an integer array?[index1, index2]?of length 2.
The tests are generated such that there is?exactly one solution. You?may not?use the same element twice.
Your solution must use only constant extra space.
Example 1:
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. We return [1, 2].
Example 2:
Input: numbers = [2,3,4], target = 6 Output: [1,3] Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].
Example 3:
Input: numbers = [-1,0], target = -1 Output: [1,2] Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].
Constraints:
2 <= numbers.length <= 3 * 104-1000 <= numbers[i] <= 1000numbers?is sorted in?non-decreasing order.-1000 <= target <= 1000- The tests are generated such that there is?exactly one solution.
class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
"""
解体思路：题目保证数组有序，且必有解
本来想到时左边枚举一个数，然后目标-当前数得值在右边二分查找
但是根据专题做的，于是想到是双指针
时间复杂度：O(1) 空间复杂度：O(1)
"""
l, r = 0, len(numbers) - 1
while l < r:
if numbers[l] + numbers[r] == target:
return [l + 1, r + 1]
elif numbers[l] + numbers[r] > target:
r -= 1
else:
l += 1
