python--leetcode496. 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.

Example 2:

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.

1. The length of both nums1 and nums2 would not exceed 1000.

这一题的意思就是给两个数组，第一个数组是第二个数组的子数组。找到第一个数组中数字在第二个数组中下一个比它大的元素，如果没有返回-1.

思路如下，无非就是遍历，判断逻辑：

class Solution(object):
    def nextGreaterElement(self, findNums, nums):
        """
        :type findNums: List[int]
        :type nums: List[int]
        :rtype: List[int]
        """
        res=[]
        for i in range(len(findNums)):
            flag=1
            max1=-9999
            for j in range(len(nums)):
                if findNums[i]==nums[j]:
                    for k in range(j,len(nums)):
                        if nums[k]>nums[j]:
                            max1=nums[k]
                            break
                    break
            if max1

三重循环，其实是比较费时间的。
介绍一种O（n）算法：

   d = {}
        st = []
        ans = []
        
        for x in nums:
            while len(st) and st[-1] < x:
                d[st.pop()] = x
            st.append(x)

        for x in findNums:
            ans.append(d.get(x, -1))
            
        return ans