Lintcode 1201. 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

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.

Example 2:

Input: [1]
Output: [-1]
Explanation: 
The number 1 can't find next greater number.

Notice

The length of given array won't exceed 10000.

Solution:

monotonic descreasing stack. 

Code (Java):

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public class Solution {     /**      * @param nums: an array      * @return: the Next Greater Number for every element      */     public int[] nextGreaterElements(int[] nums) {         // Write your code here         if (nums == null || nums.length == 0) {             return new int[0];         }                   int[] ans = new int[nums.length];                   // init to -1         for (int i = 0; i < nums.length; i++) {             ans[i] = -1;         }                   // stack is mono descreasing stack, and it stores the index         Stack<Integer> stack = new Stack<>();                   for (int i = 0; i < nums.length * 2; i++) {             int ii = i % nums.length;             while (!stack.isEmpty() && nums[ii] > nums[stack.peek()]) {                 ans[stack.pop()] = nums[ii];             }                           stack.push(ii);         }                   return ans;               } }

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class Solution {     public int[] nextGreaterElements(int[] nums) {         int[] ans = new int[nums.length];                   Stack<Integer> stack = new Stack<>();                   for (int i = nums.length * 2 - 1; i>= 0; i--) {             int ii = i % nums.length;                           while (!stack.isEmpty() && nums[ii] >= nums[stack.peek()]) {                 stack.pop();             }                           ans[ii] = stack.isEmpty() ? -1 : nums[stack.peek()];                           stack.push(ii);         }                   return ans;     } }