leetcode503. Next Greater Element II

题目要求

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.

假设如今在有一个循环数组，即数组的第一个元素是数组最后一个元素的下一个元素。如今要求生成一个新的数组，该新的数组的每个元素分别表示原数组中第一个大于该下标元素的值。

思路和代码

若是该数组不是一个循环数组，则经过栈的一轮遍历就能够将全部元素的下一个更大的元素找出来。找出来的方法即为一旦遇到一个元素大于栈顶的元素，就将栈中的元素退出，由于当前元素就是栈顶元素下一个更大的元素。

可是由于该数组是一个循环数组，因此只须要再遍历一轮数组，这一轮无需将元素入栈，只须要不断的将小于当前元素的栈顶元素弹出便可。这里要注意，数组中的最大元素在这种算法下将永远不会弹出，所以须要将结果集中的默认值设为-1.

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;
    }