503. 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 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.

首先是双层遍历的方法，复杂度比较高。代码如下：

public class Solution {    public int[] nextGreaterElements(int[] nums) {        int len = nums.length;        int max = Integer.MAX_VALUE;        int[] res = new int[len];        for (int i = 0; i < len; i ++) {            if (nums[i] >= max) {                res[i] = -1;                continue;            }            int j = 0;            for (j = 1; j < len; j ++) {                if (nums[(i + j) % len] > nums[i]) {                    res[i] = nums[(i + j) % len];                    break;                }            }            if (j == len) {                res[i] = -1;                max = Math.min(max, nums[i]);            }        }        return res;    }}

第二种方法用stack存下降的子串的index，遇到比peek()大的数就pop()填充res[stack.pop()]。代码如下：

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