leetcode503 Next Greater Element II java

Description

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.

解法

类似于Next Greater Element I

    public int[] nextGreaterElements(int[] nums) {        int n=nums.length;        int[] next = new int[n];        Arrays.fill(next,-1);        Stack<Integer> stack = new Stack<>();        for(int i=0; i<2*n; i++) {            int num = nums[i % n];            while(!stack.isEmpty() && num > nums[stack.peek()]) next[stack.pop()] = num;            if(i < n) stack.push(i);        }        return next;    }