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]
public class Solution {
public int[] nextGreaterElements(int[] nums) {
int max = Integer.MIN_VALUE;
for (int num : nums) {
max = Math.max(max, num);
}
int n = nums.length;
int[] result = new int[n];
int[] temp = new int[n * 2];
for (int i = 0; i < n * 2; i++) {
temp[i] = nums[i % n];
}
for (int i = 0; i < n; i++) {
result[i] = -1;
if (nums[i] == max) continue;
for (int j = i + 1; j < n * 2; j++) {
if (temp[j] > nums[i]) {
result[i] = temp[j];
break;
}
}
}
return result;
}
}
