You are given two arrays (without duplicates) nums1
and nums2
where nums1
’s elements are subset of nums2
. Find all the next greater numbers for nums1
's elements in the corresponding places of nums2
.
The Next Greater Number of a number x in nums1
is the first greater number to its right in nums2
. If it does not exist, output -1 for this number.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2]. Output: [-1,3,-1] Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4]. Output: [3,-1] Explanation: For number 2 in the first array, the next greater number for it in the second array is 3. For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
Note:
- All elements in
nums1
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
class Solution { public int[] nextGreaterElement(int[] nums1, int[] nums2) { if (nums1 == null || nums1.length <= 0) return new int[0]; int[] ret = new int[nums1.length]; for (int i=0; i<ret.length; i++) ret[i] = -1; for (int i=0; i<nums1.length; i++) { for (int j=0; j<nums2.length; j++) { if (nums1[i] == nums2[j]) { for (int k=j+1; k<nums2.length; k++) { if (nums2[k] > nums1[i]) { ret[i] = nums2[k]; break; } } break; } } } return ret; } }