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.

原题地址: Two Sum II - Input array is sorted

难度: Easy

思路1:

采用缓存的方式,用字典记录每一个值的索引

代码:

class Solution(object):
def twoSum(self, numbers, target):
"""
:type numbers: List[int]
:type target: int
:rtype: List[int]
"""
d = {}
for idx, num in enumerate(numbers):
if target - num in d:
idx1 = d[target-num]
return [idx1+1, idx+1]
else:
d[num] = idx

时间复杂度: O(n)

空间复杂度: O(n)

思路2:

因为数组已经排序了,可以分别在数组首尾放置一指针,逐渐向中间夹逼

代码:

class Solution(object):
def twoSum(self, numbers, target):
"""
:type numbers: List[int]
:type target: int
:rtype: List[int]
"""
i, j = 0, len(numbers)-1
while i < j:
val = numbers[i] + numbers[j]
if val == target:
return [i+1, j+1]
elif val > target:
j -= 1
else:
i += 1

时间复杂度: O(n)

空间复杂度: O(1)

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