[leetcode]Array-697. Degree of an Array

最新推荐文章于 2019-07-13 16:38:40 发布

原创最新推荐文章于 2019-07-13 16:38:40 发布 · 220 阅读

0 ·

CC 4.0 BY-SA版权

leetcode题目专栏收录该内容

10 篇文章

订阅专栏

本文介绍了一种算法，用于在给定的非空整数数组中找到具有相同度的最短连续子数组。通过示例展示了如何确定目标子数组的长度，并提供了C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

class Solution {
 2 public:
 3     int findShortestSubArray(vector<int>& nums) {
 4         int len = nums.size() ;
 5         if (len < 2)
 6             return len ;
 7         int Max = 0 ;
 8         unordered_map<int,int> startIndex , count ;
 9         int res = len ;
10         for(int i = 0 ; i < len ; i++){
11             if (startIndex.count(nums[i]) == 0)
12                 startIndex[nums[i]] = i ;
13             count[nums[i]]++ ;
14             if (count[nums[i]] > Max){
15                 res = i - startIndex[nums[i]] + 1 ;
16                 Max = count[nums[i]] ;
17             }
18             if (count[nums[i]] == Max){
19                 res = min(i - startIndex[nums[i]] + 1 , res) ;  
20             }
21         }
22         return res ;
23     }
24 };