Leetcode: Next Greater Element II

Posted on February 20, 2018September 23, 2019 by braindenny

Next Greater Element II

Similar Problems:

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.

// https://code.dennyzhang.com/next-greater-element-ii
// Basic Ideas: monotone stack
//
//   1 3 2 1 4
//   1 3 2 1 4 1 3 2 1 4
// Complexity: Time O(n), Space O(n)
func nextGreaterElements(nums []int) []int {
    nums2 := make([]int, len(nums)*2)
    for i, v := range nums {
        nums2[i] = v
        nums2[i+len(nums)] = v
    }
    // monotone stack
    l := make([]int, len(nums2))
    stack := []int{}
    for i, _ := range l {
        l[i] = -1
    }
    for i, v := range nums2 {
        for len(stack)>0 && nums2[stack[len(stack)-1]] < v {
            l[stack[len(stack)-1]] = v
            stack = stack[0:len(stack)-1]
        }
        stack = append(stack, i)
    }
    return l[0:len(nums)]
}

## https://code.dennyzhang.com/next-greater-element-ii
## Basic Ideas: 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

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