[LeetCode] 496. Next Greater Element I_Easy tag: Stack
2018-08-18 05:44 Johnson_强生仔仔 阅读(184) 评论(0) 编辑 收藏 举报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.
这个题目就是用stack, 分别将nums2里面每个对应的next bigger number找到, 如果没有, 那么为-1, 并且存到d里面. T :O(m) m = len(nums)
Code
class Solution(object): def nextGreaterElement(self, findNums, nums): stack,d, ans = [],{}, [] for num in nums: while stack and num > stack[-1]: d[stack.pop()] = num stack.append(num) while stack: d[stack.pop()] = -1 for num in findNums: ans.append(d[num]) return ans