Description:
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.
Solution
class Solution { public int[] nextGreaterElement(int[] nums1, int[] nums2) { if(nums1==null||nums2 == null ){ return null; } int [] result = new int[nums1.length]; List<Integer> intList = new ArrayList<Integer>(); for (int i : nums2) { intList.add(i); } for(int i = 0; i< nums1.length; i++){ for(int j = intList.indexOf(nums1[i]); j< nums2.length; j++ ){ if(j ==-1){ result[i] = -1; break; } if(nums2[j]>nums1[i]){ result[i] = nums2[j]; break; } result[i] =-1; } } return result; } }