问题
给定一个非空且只包含非负数的整数数组 nums, 数组的度的定义是指数组里任一元素出现频数的最大值。
你的任务是找到与 nums 拥有相同大小的度的最短连续子数组，返回其长度。
输入: [1, 2, 2, 3, 1]
输出: 2
解释:
输入数组的度是2，因为元素1和2的出现频数最大，均为2.
连续子数组里面拥有相同度的有如下所示:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
最短连续子数组[2, 2]的长度为2，所以返回2.
输入: [1,2,2,3,1,4,2]
输出: 6
注意:
nums.length 在1到50,000区间范围内。
nums[i] 是一个在0到49,999范围内的整数。
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.
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.
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.
示例
public class Statistics {
//统计频次
public int Count { get; set; }
//开始索引
public int StartIndex { get; set; }
//结束索引
public int EndIndex { get; set; }
}
public class Program {
public static void Main(string[] args) {
int[] nums = null;
nums = new int[] { 1, 2, 2, 3, 1 };
var res = FindShortestSubArray(nums);
Console.WriteLine(res);
Console.ReadKey();
}
private static int FindShortestSubArray(int[] nums) {
//哈希存放统计数据
var dic = new Dictionary<int, Statistics>();
for(int i = 0; i < nums.Length; i++) {
//如果有键，频次+1，记录结束索引
if(dic.ContainsKey(nums[i])) {
dic[nums[i]].Count++;
dic[nums[i]].EndIndex = i;
} else {
//如果没有键，赋初始值
dic.Add(nums[i], new Statistics {
Count = 1,
StartIndex = i,
EndIndex = i
});
}
}
//降序排列
var order = dic.OrderByDescending(s => s.Value.Count).ToList();
//记录数组的度
var maxCount = order[0].Value.Count;
//记录最小长度
var minLength = int.MaxValue;
//先筛选出所有与数组的度相同的统计结果
order.Where(s => s.Value.Count == maxCount)
//转换成List
.ToList()
//遍历所有List中的数据找出最小Length
.ForEach(s => minLength = Math.Min(
minLength,
s.Value.EndIndex - s.Value.StartIndex + 1
));
//返回最小Length
return minLength;
}
}以上给出1种算法实现，以下是这个案例的输出结果：
2分析：
以上解法在最坏的情况下，var order = dic.OrderByDescending(s => s.Value.Count).ToList() 执行所消耗的时间最多，因为原数组可能没有重复的元素。若OrderByDescending基于比较的先进算法，那么以上解法的时间复杂度为： ；若基于空间换时间的基数、计数或桶排序，那么以上解法的时间复杂度为：
。