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.

Tips:

stack: O(n)

1. 先从倒数第二个数开始，创建一个递减栈，把栈内比当前数小的数都pop后push当前数。
2. 从最后一个数开始，在递减栈中将比自己小的数都扔掉，创建新的递减栈。因为栈先入后出，所以第二遍等于从头开始找比自己大的第一个数。此时res[i] = stack.peek()，然后再将当前数入栈创建新的递减栈。

brute forth: O(n^2)

Code:

stack: 屌得飞

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

brute forth:

public int[] nextGreaterElements2(int[] nums) {
    int  n = nums.length;
    int[] res = new int[n];
    for (int i = 0; i < n; i++) {
        res[i] = -1;
        for (int k = i+1; k < i + n; k++) {
            if (nums[k%n] > nums[i]){
                res[i] = nums[k%n];
                break;
            }
        }
    }
    return res;
}