Leetcode #167. Two Sum II - Input array is sorted
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. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
在一个sorted list中寻找两数之和等于给定的target, 一般用双指针的方式解决。
1 class Solution(object): 2 def twoSum(self, numbers, target): 3 """ 4 :type numbers: List[int] 5 :type target: int 6 :rtype: List[int] 7 """ 8 9 n = len(numbers) 10 i = 0 11 j = n-1 12 13 while i < j: 14 if numbers[i] + numbers[j] == target: 15 return [i+1,j+1] 16 elif numbers[i] + numbers[j] < target: 17 i += 1 18 elif numbers[i] + numbers[j] > target: 19 j -= 1
如果使用binary search的方法，其实就是固定一个数 numbers[i]，然后从它后面的数中寻找 target - numbers[i]。 可以得到一个O(nlogn)的解法。
1 class Solution(object): 2 def twoSum(self, numbers, target): 3 """ 4 :type numbers: List[int] 5 :type target: int 6 :rtype: List[int] 7 """ 8 if not numbers or len(numbers) == 0: 9 return [] 10 11 n = len(numbers) 12 13 for i in range(n): 14 left, right = i + 1, n-1 15 while left <= right: 16 mid = left + (right - left)/2 17 if numbers[mid] == target - numbers[i]: 18 return [i+1, mid+1] 19 elif numbers[mid] < target - numbers[i]: 20 left = mid + 1 21 else: 22 right = mid - 1 23 return []
