You are given two arrays (without duplicates) nums1 and nums2 where nums1โs 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.
Time complexity : O(m+n). The entire nums array(of size n) is scanned only once. The stack's n elements are popped only once. The findNums array is also scanned only once.
Space complexity : O(m+n). stack and map of size n is used. res array of size mm is used, where nn refers to the length of the nums array and mm refers to the length of the findNums array.
classSolution {publicint[] nextGeneraterElement(int[] findNums,int[] nums) {Stack<Integer> stack =newStack<>();HashMap<Integer,Integer> map =newHashMap<>();int[] res =newint[findNums.length];for (int i =0; i <nums.length; i++) {// ๅ่ฐ้ๅๆ while (!stack.empty() && nums[i] >stack.peek()) {int small =stack.pop();map.put(small, nums[i]); }stack.push(nums[i]); }while (!stack.empty()) {map.put(stack.pop(),-1); }for (int i =0; i <findNums.length; i++) { res[i] =map.get(findNums[i]); }return res; }}