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.

Solution

Stack, time O(n), space O(n)

这道题蛮有意思的，如果用array去做需要O(n ^ 2)的时间，但是用stack做又没什么头绪，后来发现可以换一种思路去用stack。往常某元素x入栈的时候，x可能成为栈中元素的next greater value，但是我们换一个思路，栈中元素储存next greater value candidates，某元素x入栈的时候则在栈中寻找它的next greater value。

由于是cycler array，我们需要将array中元素逆序入栈，然后逆序遍历array，这样遍历到元素x的时候，能够保证x的cycler next values都在栈中。

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        Stack<Integer> stack = new Stack<>();
        int n = nums.length;
        int[] res = new int[n];
        Arrays.fill(res, -1);
        
        for (int i = n - 1; i >= 0; --i) {
            stack.push(nums[i]);
        }
        
        for (int i = n - 1; i >= 0; --i) {
            while (!stack.empty() && stack.peek() <= nums[i]) {
                // pop directly because nums[i] is better than stack.peek()
                // "greater" means nums[i] is larger and i is smaller
                stack.pop();    
            }
            if (!stack.empty()) {
                res[i] = stack.peek();
            }
            stack.push(nums[i]);
        }
        
        return res;
    }
}