1. Description
2. Solution
**解析:**Version 1,由于元素是唯一的,通过循环找出每个nums2
中的满足条件结果保存到字典中,遍历nums1
,获得结果。Version 2通过使用栈来寻找满足条件的结果,减少搜索时间。
- Version 1
class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
stat = {}
for i in range(len(nums2)):
for j in range(i 1, len(nums2)):
if nums2[j] > nums2[i]:
stat[nums2[i]] = nums2[j]
break
if nums2[i] not in stat:
stat[nums2[i]] = -1
result = [stat[x] for x in nums1]
return result
- Version 2
class Solution:
def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
stat = {num: -1 for num in nums2}
stack = []
for num in nums2:
while stack and stack[-1] < num:
stat[stack.pop()] = num
stack.append(num)
result = [stat[x] for x in nums1]
return result
Reference
- https://leetcode.com/problems/next-greater-element-i/