LeetCode 496. Next Greater Element I
原题链接在这里:https://leetcode.com/problems/next-greater-element-i/description/
题目:
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.
题解:
对于nums2中的每一个元素,找到next greater element并对应放到HashMap<Integer, Integer> hm 中.
如何一遍找到呢,通过Stack<Integer> stk, 从右向左扫面,新的元素num会把stk顶部比num小的全部pop出去, 剩下的顶部就是next greater element, 存到hm中.
stk便成了从顶向下增大的栈. 最后利用hm把findNums的每个元素对应的next greater element找出来.
Time Complexity: O(n), n = nums2.length.
Space: O(n).
AC Java:
1 class Solution { 2 public int[] nextGreaterElement(int[] nums1, int[] nums2) { 3 HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>(); 4 Stack<Integer> stk = new Stack<Integer>(); 5 6 for(int i = nums2.length-1; i>=0; i--){ 7 while(!stk.isEmpty() && nums2[i]>=nums2[stk.peek()]){ 8 stk.pop(); 9 } 10 11 int nextGreater = stk.isEmpty() ? -1 : nums2[stk.peek()]; 12 hm.put(nums2[i], nextGreater); 13 14 stk.push(i); 15 } 16 17 int [] res = new int[nums1.length]; 18 for(int i = 0; i<nums1.length; i++){ 19 res[i] = hm.get(nums1[i]); 20 } 21 22 return res; 23 } 24 }
We could also iterate nums1 from left to right.
When hitting a greater element x, pop up all the smaller element in the stack and mark their next greater element as x and store in the map.
Time Complexity: O(n). n = nums1.length.
Space: O(n).
AC Java:
1 class Solution { 2 public int[] nextGreaterElement(int[] nums1, int[] nums2) { 3 HashMap<Integer, Integer> hm = new HashMap<>(); 4 Stack<Integer> stk = new Stack<>(); 5 for(int num : nums2){ 6 while(!stk.isEmpty() && stk.peek() < num){ 7 hm.put(stk.pop(), num); 8 } 9 10 stk.push(num); 11 } 12 13 int n = nums1.length; 14 int [] res = new int[n]; 15 for(int i = 0; i < n; i++){ 16 res[i] = hm.getOrDefault(nums1[i], -1); 17 } 18 19 return res; 20 } 21 }
AC C++:
1 class Solution { 2 public: 3 vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) { 4 unordered_map<int , int> map; 5 stack<int> stk; 6 for(int num : nums2){ 7 while(!stk.empty() && stk.top() < num){ 8 map[stk.top()] = num; 9 stk.pop(); 10 } 11 12 stk.push(num); 13 } 14 15 vector<int> res(nums1.size(), -1); 16 for(int i = 0; i < res.size(); i++){ 17 auto itr = map.find(nums1[i]); 18 if(itr != map.end()){ 19 res[i] = itr->second; 20 } 21 } 22 23 return res; 24 } 25 };
AC Python:
1 class Solution: 2 def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]: 3 hm = {} 4 stk = [] 5 6 for num in nums2: 7 while stk and stk[-1] < num: 8 hm[stk.pop()] = num 9 stk.append(num) 10 11 return [hm.get(x, -1) for x in nums1]
跟上Next Greater Element II, Daily Temperatures, Minimum Cost Tree From Leaf Values, Online Stock Span, Number of Visible People in a Queue.