動機

Monotonic Stack其中一個功能就是在pop的時候可以知道下一個最大

Problem

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, return -1 for this number.

 

Example 1:

Input: nums = [1,2,1]Output: [2,-1,2]Explanation: The first 1's next greater number is 2; The number 2 can't find next greater number. The second 1's next greater number needs to search circularly, which is also 2.

Example 2:

Input: nums = [1,2,3,4,3]Output: [2,3,4,-1,4]

 

Constraints:

  • 1 <= nums.length <= 104
  • -109 <= nums[i] <= 109

Sol

circular? 長度兩倍就好

class Solution:
    def nextGreaterElements(self, nums: List[int]) -> List[int]:
        stk = []
        ret = []
        nums += nums
        for i,n in enumerate(nums):
            while stk and nums[stk[-1]] < n:
                ret[stk.pop()] = n
            stk += i,
            ret += -1,
        return ret[:len(nums)//2]