python leetcode 496. Next Greater Element I

最新推荐文章于 2022-01-30 21:29:27 发布

原创最新推荐文章于 2022-01-30 21:29:27 发布 · 808 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#python #leetcode

leetcode_python 专栏收录该内容

71 篇文章

订阅专栏

本文介绍了一个算法问题，即如何在两个数组中找到特定元素的下一个更大元素。具体来说，对于数组nums1中的每个元素，在数组nums2中找到该元素右侧的第一个更大的元素。如果不存在这样的元素，则返回-1。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

找到nums2的子集nums1中每个元素在nums2中的对应位置的右侧第一个比自身大的元素，若不存在，则返回-1。

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.

class Solution:
def nextGreaterElement(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
ans = []
for i in range(len(nums1)):
start = nums2.index(nums1[i])
nums3 = nums2[start :]
for j in range(len(nums3)):
if nums1[i] < nums3[j]:
ans.append(nums3[j])
break
else:
ans.append(-1)
return ans