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:

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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:

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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:

  1. All elements in nums1 and nums2 are unique.
  2. The length of both nums1 and nums2 would not exceed 1000.

思路

利用栈的方法，可以满足题意。

因为如果一个数组的元素是递减的，这些递减的元素都不可能互为“Next Greater Element”。直到有一个递增的元素，才能满足题意。我们拿到这个递增的元素，不断地和栈顶的元素进行比较，直到该元素比栈顶元素小为止。出栈的这些元素的“Next Greater Element”，就是这个递增的元素的值。

由于没有重复，可以用map来一一对应。

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class Solution(object):
    def nextGreaterElement(self, findNums, nums):
        """
        :type findNums: List[int]
        :type nums: List[int]
        :rtype: List[int]
        """
        resultmap = {}
        tmpelem = [-1]
        while (nums):
            tmp = nums.pop()
            if (nums):
                if (nums[-1] < tmp):
                    tmpelem.append(tmp)
            
            resultmap[tmp] = -1
            for i in tmpelem:
                if (tmp < i):
                    resultmap[tmp] = i
        
        result = []
        for i in findNums:
            result.append(resultmap[i])
            
        return result