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.
solution
Same as 496-next-greater-element-i, but we iterate through the array twice using modulo.
def nextGreaterElements(self, nums: List[int]) -> List[int]:
stack = []
n = len(nums)
ans = [-1] * n
"""
[5,4,3,2,1] -> [5,4,3,2,1,5,4,3,2]
"""
for i in range(2*n-1):
while stack and stack[-1][0] < nums[i%n]:
_popped, idx = stack.pop()
ans[idx] = nums[i%n]
stack.append((nums[i%n], i%n))
return ans