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## Problem

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.

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, return -1 for this number.

Example 1:

Input: nums = [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.


Example 2:

Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]


Constraints:

• 1 <= nums.length <= 10^4
• -10^9 <= nums[i] <= 10^9

## Code

496

class Solution {
public int[] nextGreaterElements(int[] nums) {
int len = nums.length;
int[] res = new int[len];
Stack<Integer> stack = new Stack<>();

Arrays.fill(res, -1);

// 针对循环数组, 可以把数组遍历两边
for (int i = 0; i < 2 * len; i++) {
int num = nums[i % len];

while (!stack.isEmpty() && nums[stack.peek()] < num) {
res[stack.pop()] = num;
}

// 只在第一轮遍历时放入index
if (i < len) {
stack.push(i);
}
}

return res;
}
}