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.
我们可以使用栈来进行优化上面的算法，我们遍历两倍的数组，然后还是坐标i对n取余，取出数字，如果此时栈不为空，且栈顶元素小于当前数字，说明当前数字就是栈顶元素的右边第一个较大数，那么建立二者的映射，并且去除当前栈顶元素，最后如果i小于n，则把i压入栈。因为res的长度必须是n，超过n的部分我们只是为了给之前栈中的数字找较大值，所以不能压入栈，参见代码如下：
没有用hashMap 存-1的值, 用的是Arrays.fill(nums, -1)
用到了 stack 来求原序列中不打乱相对次序的递减子序列的技巧:
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;
}