[LeetCode] 167. Two Sum II – Input Array Is Sorted

https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/

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].

Github Solution

#time: O(n), space: O(1)
class Solution(object):
    def twoSum(self, numbers, target):
        if not numbers or len(numbers) < 2:
            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 []

Solution:

1. instead of using a hash map, we use two pointers: left_pointer and right_pointer, pointing to the leftmost value and the rightmost value.
2. left_pointer points to the next value if the sum of two pointers is smaller than the target value
3. right_pointer points to the last value if the sum of two pointers is larger than the target value
4. return their index if the sum of two pointers equal to the target value

解法：

1. 不用hash, 直接用指向最左和最右兩個pointer, left_pointer, right_pointer
2. 左和右兩個pointer所指的值相加，比target value小的時候，left_pointer++
3. 左和右兩個pointer所指的值相加，比target value大的時候，right_pointer–
4. 如果左和右兩個pointer所指的值相加，return 這兩個pointer的index值