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
Solution (Java)
class Solution {
public int[] nextGreaterElements(int[] nums) {
int[] result = new int[nums.length];
Deque<Integer> stack = new ArrayDeque<>();
for (int i = nums.length * 2 - 1; i >= 0; i--) {
while (!stack.isEmpty() && nums[stack.peek()] <= nums[i % nums.length]) {
stack.pop();
}
result[i % nums.length] = stack.isEmpty() ? -1 : nums[stack.peek()];
stack.push(i % nums.length);
}
return result;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).