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.

假设现在在有一个循环数组，即数组的第一个元素是数组最后一个元素的下一个元素。现在要求生成一个新的数组，该新的数组的每一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

如果该数组不是一个循环数组，则通过栈的一轮遍历就可以将所有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，因为当前元素就是栈顶元素下一个更大的元素。

但是因为该数组是一个循环数组，所以只需要再遍历一轮数组，这一轮无需将元素入栈，只需要不断的将小于当前元素的栈顶元素弹出即可。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，因此需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;
    }