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.
Note: All elements in nums1 and nums2 are unique. The length of both nums1 and nums2 would not exceed 1000.
給出兩個不含重復元素的array nums1, nums2。 nums1 是nums2的子集（nums2包含nunms1的所有元素）。
Next Greater Element 指的是找出在nums1里面的每個元素在nums2中相同元素的右邊比它本身大的元素， 如果沒有或者比其小則為-1。
class Solution(object): def nextGreaterElement(self, findNums, nums): """ :type findNums: List[int] :type nums: List[int] :rtype: List[int] """ result = [] for c in findNums: pos = nums.index(c) while pos <= (len(nums) - 1): if pos == len(nums) - 1: result.append(-1) break else: if nums[pos+1] > c: result.append(nums[pos+1]) break else: pos += 1 return result附上leeetcode答案 O(n)：
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新聞熱點
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