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.

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.
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.

首先是自己的想法，比较笨拙，就是找到相同的数后驱查找后面比它大的数，找到就终止循环，将数字记入，找不到就记入-1.

    def nextGreaterElement(self, findNums, nums):
        """
        :type findNums: List[int]
        :type nums: List[int]
        :rtype: List[int]
        """
        gle = []
        for i in range(len(findNums)):
            flag = False
            for j in range(len(nums)):
                if findNums[i] == nums[j]:
                    flag = True
                if (flag and nums[j] > findNums[i]):
                        gle.append(nums[j])
                        flag = False
                        break
            if flag:
                gle.append(-1)
        return gle

在网上看其他的方法，用stack的想法：

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

方法如下,用一个字典记录，记录比当前这个数大的后面的第一个数：

    def nextGreaterElement(self, findNums, nums):
        """
        :type findNums: List[int]
        :type nums: List[int]
        :rtype: List[int]
        """
        m = {}
        s = []
        res = []
        for i in nums:
            while len(s) and s[-1]<i:
                m[s.pop()] = i
            s.append(i)
        for j in findNums:
            res.append(m.get(j,-1))
        return res