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.
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; }