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.
一刷
题解：与496不一样，数组变成了首位相连的数组。于是for-loop的范围变成了2*len. 方法与496一样。
public class Solution {
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;
}
}
