原题链接https://leetcode.com/problems/next-greater-element-ii/
Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. 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, output -1 for this number.
Example 1:
Input: [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.
这里要输出的是循环数组中比当前数第一个大的值.
class Solution:
def nextGreaterElements(self, nums: List[int]) -> List[int]:
n = len(nums)
nums = nums * 2
stack = []
res = [-1] * 2 * n
for i, t in enumerate(nums):
while stack and nums[stack[-1]] < t:
res[stack[-1]] = t
stack.pop()
stack.append(i)
return res[:n]
