7/20 每日一題(兩個數組 數組2是數組1的子數組,返回數組1元素在數組2中後面項次中比他更大者)

 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.

 

Follow up: Could you find an O(nums1.length + nums2.length) solution?

-----------不動腦方法 不滿足時間複雜度要求

class Solution:

    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:

        res=[]

        n=0

        while n < len(nums1):

            element=nums1[n]

            for i in nums2[nums2.index(element):]:

                ele=-1

                if i > element :

                    ele=i #將比他大的值賦予到ele中

                    break #跳出這該死的迴圈

            res.append(ele)

            n+=1 #繼續檢查下一項

        return res

----------------------使用堆疊

class Solution:

    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:

        res=[]#結果

        stack=[]#堆疊

        bigger_dict={}

        for i in nums2:

            while stack and i > stack[-1]:

                #當前元素比堆疊頂部大的時候記錄到字典,key是之前推入堆疊的元素value是目前更大的元素

                bigger_dict[stack.pop()]=i

           

            stack.append(i)

        for i in nums1:

            res.append(bigger_dict.get(i,-1))

        return res