# LeetCode - 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.

class Solution {
public int[] nextGreaterElements(int[] nums) {
if (nums == null || nums.length <= 0)
return new int[0];
int[] ret = new int[nums.length];
for (int i=0; i<ret.length; i++)
ret[i] = -1;
for (int i=0; i<nums.length; i++) {
int flag = 0;
for (int j=i+1; j<nums.length; j++) {
if (nums[j] > nums[i]) {
ret[i] = nums[j];
flag = 1;
break;
}
}
if (flag == 0) {
for (int k=0; k<i; k++) {
if (nums[k] > nums[i]) {
ret[i] = nums[k];
break;
}
}
}
}
return ret;
}
}

posted @ 2018-08-16 14:58  Pickle  阅读(168)  评论(0编辑  收藏  举报