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:
All elements in nums1 and nums2 are unique.
The length of both nums1 and nums2 would not exceed 1000.
这题描述的需求是：
给我们两个数组 比如 nums1 = [1，2，3] nums2 = [1，2，3，4，5，6]
需要我们求出的结果也是一个数组，这个数组，里面的数值一次是：
对num1里面每一个数字x ，找到在num2里x出现的右侧最近的一个比x大的数字。
如过nums里x的右侧没有比x大的 就用-1代表这个数字的结果
比如：
Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
4在num2中的右侧没有比4大的 所以结果是-1
1在num2中 右侧最近的比他大的是3
2在num2中没有右侧数字了 结果是-1
我的python代码：
1 class Solution(object):
2 def nextGreaterElement(self, findNums, nums):
3 """
4 :type findNums: List[int]
5 :type nums: List[int]
6 :rtype: List[int]
7 """
8 res = []
9 for i in findNums:
10 index = nums.index(i)
11 index2 = -1
12 for j in range(index + 1, len(nums)):
13 if nums[j] > i:
14 index2 = j
15 break
16 if index2 == -1:
17 res.append(-1)
18 else :
19 res.append( nums[index2])
20 return res
21
22
23
24
25 if __name__ == '__main__':
26 s = Solution()
27 res = s.nextGreaterElement([4,1,2], [1,3,4,2] )
28 print(res)