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描述
You are given two integer arrays nums1 and nums2 both of unique elements, where nums1 is a 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, return -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 <= nums1.length <= nums2.length <= 1000
0 <= nums1[i], nums2[i] <= 10^4
All integers in nums1 and nums2 are unique.
All the integers of nums1 also appear in nums2.
解析
根据题意，num1 是 num2 的子集，找出 num1 中每个元素在 num2 中的对应位置之后的第一个大于它的数，如果没有则结果为 -1 。我这里定义了一个方法 find ，就是为了在 num2 中的 idx 位置之后找比 nums2[idx] 大的数字。直接遍历 num1 ，得到元素在 num2 中的索引，然后使用我定义的 find 函数，得到的结果追加到 result 中，遍历结束即可得到结果。
解答
class Solution(object):
def nextGreaterElement(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
def find(A, idx):
result = -1
for i in range(idx, len(A)):
if A[i]>A[idx]:
result = A[i]
break
return result
result = []
for n in nums1:
idx = nums2.index(n)
result.append(find(nums2, idx))
return result
运行结果
Runtime: 80 ms, faster than 21.98% of Python online submissions for Next Greater Element I.
Memory Usage: 13.3 MB, less than 100.00% of Python online submissions for Next Greater Element I.
解析
上面是按照题意解答的一种思路，比较简单，还有一种方法利用了字典和栈，因为每个 nums1 中的数字都在 nums2 中，遍历 nums2 中的每个数字，找出每个元素及其后面第一个大的元素存入字典 d 中，然后遍历 num1 中的每个数字，从 d 中直接获取值，如果没有则直接返回 -1 。
解答
class Solution(object):
def nextGreaterElement(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
result = []
stack = []
d = {}
for num in nums2:
while len(stack) and (stack[-1]<num):
d[stack.pop()] = num
stack.append(num)
for num in nums1:
result.append(d.get(num,-1))
return result
运行结果
Runtime: 28 ms, faster than 96.30% of Python online submissions for Next Greater Element I.
Memory Usage: 13.8 MB, less than 33.55% of Python online submissions for Next Greater Element I.
原题链接：leetcode.com/problems/ne…
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