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
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
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o(n^2)的算法还是很好想的 是被注释掉的那部分
看见题目标签说stack
然后看了给的solution,很精致的用stack的方法
思路如下
Key observation:
Suppose we have a decreasing sequence followed by a greater number
For example [5, 4, 3, 2, 1, 6]
then the greater number 6
is the next greater element for all previous numbers in the sequence
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We use a stack to keep a
decreasing sub-sequence, whenever we see a number
For example
x
greater than
stack.peek()
we pop all elements less than
x
and for all the popped ones, their next greater element is
x
For example
[9, 8, 7, 3, 2, 1, 6]
The stack will first contain
[9, 8, 7, 3, 2, 1]
and then we see
6
which is greater than
1
so we pop
1 2 3
whose next greater element should be
6
class Solution(object):
def nextGreaterElement(self, findNums, nums):
a = findNums
b = nums
m = {}
s = []
for i in b:
while len(s) and s[-1] < i:
m[s.pop()] = i
s += i,
res = []
for i in a:
res += m.get(i,-1),
return res
"""
#a.sort()
#b.sort()
#print a,b
res = []
for i in a:
pos = b.index(i)
#bb = b[pos:]
j = pos
while j < len(b):
if b[j] > i:
res += b[j],
break
j += 1
if j == len(b):
res += -1,
return res
:type findNums: List[int]
:type nums: List[int]
:rtype: List[int]
"""