Next Greater Element
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.
1 public class Solution { 2 public int[] nextGreaterElement(int[] findNums, int[] nums) { 3 Stack<Integer> stk = new Stack<Integer>(); 4 HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); 5 6 int n = nums.length; 7 for(int i = n - 1; i >= 0; i--) { 8 while (!stk.empty() && nums[i] >= stk.peek()) 9 stk.pop(); 10 11 map.put(nums[i], stk.empty() ? -1 : stk.peek()); 12 stk.push(nums[i]); 13 } 14 15 int m = findNums.length; 16 int[] result = new int[m]; 17 for (int i = 0; i < m; i++) { 18 result[i] = map.get(findNums[i]); 19 } 20 21 return result; 22 } 23 }