LeetCode 496. 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.

这道题是要求一个数组里的元素在另一个数组里，下一个比它大的元素是多少。最暴力的办法就是双重for loop，在另一个数组里找这个元素在哪儿然后看看后面哪个比它大。O(n^2)复杂度。这题的巧妙解法叫做monotonic stack，就是维护一个递减序列的stack，保证每次往stack里push元素的时候，这个元素都是当前序列下最小的。而如果要push的元素比当前的stack top要小，就把stack top pop出来，直到剩下的元素都比要push的元素大。被pop出来的这些元素，它的next greater element就是这个要push的元素。最后stack里剩下的元素就是没有next greater element的元素。复习了一下Java stack的用法，还需要一个map来记录在pop时每个数字对应的next greater element。代码写起来没什么坑，主要是思路难想。

Runtime: 10 ms, faster than 28.32% of Java online submissions for Next Greater Element I.

Memory Usage: 44.5 MB, less than 25.83% of Java online submissions for Next Greater Element I.

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