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]