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
SOLUTION:
class Solution:
def nextGreaterElements(self, nums: List[int]) -> List[int]:
n = len(nums)
op = []
for i in range(n):
found = False
for j in range(i + 1, i + n + 1):
if nums[j % n] > nums[i]:
op.append(nums[j % n])
found = True
break
if not found:
op.append(-1)
return op
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