503. Next Greater Element II

503. Next Greater Element II

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.

Note: The length of given array won’t exceed 10000.

时间复杂度O(n ^ 2)。

class Solution { public int[] nextGreaterElements(int[] nums) { int size = nums.length; int[] ans = new int[size]; Arrays.fill(ans, -1); for (int i = 0; i < size; i++) { for (int j = i + 1; j % size != i; j++) { if (nums[j % size] > nums[i]) { ans[i] = nums[j % size]; break; } } } return

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。