[LeetCode] 697. Degree of an Array_Easy tag: Hash Table
2018-08-13 05:46 Johnson_强生仔仔 阅读(138) 评论(0) 编辑 收藏 举报Given a non-empty array of non-negative integers nums
, the degree of this array is defined as the maximum frequency of any one of its elements.
Your task is to find the smallest possible length of a (contiguous) subarray of nums
, that has the same degree as nums
.
Example 1:
Input: [1, 2, 2, 3, 1] Output: 2 Explanation: The input array has a degree of 2 because both elements 1 and 2 appear twice. Of the subarrays that have the same degree: [1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2] The shortest length is 2. So return 2.
Example 2:
Input: [1,2,2,3,1,4,2] Output: 6
Note:
nums.length
will be between 1 and 50,000.nums[i]
will be an integer between 0 and 49,999.
这个题目思路就是用Counter去计算数量, 然后用d1, d2 分别存每个element存在的最左端和最右端, 然后min(ans, d2[k] - d1[k] + 1), 这里ans初始化为len(ans).
Code
class Solution: def degreeArray(self, nums): d = collections.Counter(nums) fre = max(d.values()) d1, d2, ans = {}, {}, len(nums) for index, num in enumerate(nums): if num not in d1: d1[num] = index d2[num] = index for k in d.keys(): if d[k] == fre: ans = min(ans, d2[k] - d1[k] + 1) return ans