LeetCode 496. Next Greater Element I
496. Next Greater Element I
Description Submission Solutions
- Total Accepted: 6642
- Total Submissions: 11247
- Difficulty: Easy
- Contributors: love_FDU_llp
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.
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【题目分析】
给定两个数组num1和num2,两个数组中的元素都是唯一的并且num1是num2的一个子集。求num1中的元素对应的num2中相同的元素在num2中右边第一个大于它的元素。
【思路】
一开始就没有什么想法。大神的回答如下:
Key observation:
Suppose we have a decreasing sequence followed by a greater number
For example [5, 4, 3, 2, 1, 6]
then the greater number 6
is the next greater element for all previous numbers in the sequence
We use a stack to keep a decreasing sub-sequence, whenever we see a number x
greater than stack.peek()
we pop all elements less than x
and for all the popped ones, their next greater element is x
For example [9, 8, 7, 3, 2, 1, 6]
The stack will first contain [9, 8, 7, 3, 2, 1]
and then we see 6
which is greater than 1
so we pop 1 2 3
whose next greater element should be 6。
【java代码】
1 public int[] nextGreaterElement(int[] findNums, int[] nums) { 2 Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x 3 Stack<Integer> stack = new Stack<>(); 4 for (int num : nums) { 5 while (!stack.isEmpty() && stack.peek() < num) 6 map.put(stack.pop(), num); 7 stack.push(num); 8 } 9 for (int i = 0; i < findNums.length; i++) 10 findNums[i] = map.getOrDefault(findNums[i], -1); 11 return findNums; 12 }