Next Greater Element I

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.

思路：这个题目很好，是单调栈的经典题型；找右边第一个比 num[i]要大的数，那么从右边来的比当前num 小的数，就没必要存，stack里面只存比stack peek 大的数，来的数比peak大，那么踢掉peek，同时踢掉的时候记录peek的next great elment就是踢掉的那个值，整个stack就是一个单调递减栈。这题精妙的地方在于,用hashmap去存右边第一个比他大的数的mapping，也就是存右边第一个踢到这个数的num；如果没有人踢他，代表右边没有人比他大；getOrDefault(nums1[i], -1);

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int[] res = new int[n];
        Arrays.fill(res, -1);
        HashMap<Integer, Integer> hashmap = new HashMap<>();
        Stack<Integer> stack = new Stack<>();
        for(int i = 0; i < nums2.length; i++) {
            if(stack.isEmpty()) {
                stack.push(i);
            } else {
                while(!stack.isEmpty() && nums2[stack.peek()] < nums2[i]) {
                    int index = stack.pop();
                    hashmap.put(nums2[index], nums2[i]);
                }
                stack.push(i);
            }
        }
        for(int i = 0; i < n; i++) {
            if(hashmap.containsKey(nums1[i])) {
                res[i] = hashmap.get(nums1[i]);
            }
        }
        return res;
    }
}