Next Greater Element II

Similar Problems:

- Leetcode: Leetcode: Next Greater Element I
- Review: Monotone Stack Or Monotone Queue Problems, Tag: monotone

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.

Note: The length of given array won’t exceed 10000.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/next-greater-element-ii ## Basic Idea: Descending stack ## ## Complexity: Time O(n), Space O(n) class Solution(object): def nextGreaterElements(self, nums): """ :type nums: List[int] :rtype: List[int] """ length = len(nums) stack = [] res = [None]*length for i in xrange(length*2): i = i % length # if current element is bigger, it's the target of previous undecided elements while len(stack) != 0 and nums[i] > nums[stack[-1]]: k = stack.pop() res[k] = nums[i] # No need to check if already examined if res[i] is None: stack.append(i) for i in xrange(length): if res[i] is None: res[i] = -1 return res

Share It, If You Like It.