496. Next Greater Element I (E)

https://leetcode.com/problems/next-greater-element-i/

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.

Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.

Constraints:

• 1 <= nums1.length <= nums2.length <= 1000

• 0 <= nums1[i], nums2[i] <= 104

• All integers in nums1 and nums2 are unique.

• All the integers of nums1 also appear in nums2.

Follow up: Could you find an O(nums1.length + nums2.length) solution?

Solution:

单调栈：

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        
        int[] result = new int[nums1.length];
        Map<Integer, Integer> map= new HashMap<Integer, Integer>();
        for(int i = 0;i< nums1.length; i++)
        {
            map.put(nums1[i], i);
        }
        
        int[] nums2Result = new int[nums2.length];
        Stack<Integer> stack = new Stack<Integer>();
        
        for(int i = nums2.length-1; i >= 0; i--)
        {
            while(!stack.empty() && stack.peek() <= nums2[i])
            {
                stack.pop();
            }
            nums2Result[i] = stack.empty() ? -1: stack.peek();
            if(map.containsKey(nums2[i]))
            {
                result[map.get(nums2[i])] = nums2Result[i];
            }
            stack.push(nums2[i]);
        }
        return result;
        
    }
}