LeetCode.697 Degree of an Array

题目：

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: 2Explanation: 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.

分析：

class Solution {    public int findShortestSubArray(int[] nums) {        //给定非空数组，其中字串中的度的定义：数组中出现频率最高的数字        //找出其中最小的连续子串，该子串包含度的任何一个数字（即必须至少有一种度的所有元素在其中）        //思路：第一步：找出度的数。第二步，超出每个度的起始位置。最后起始位置最小的便是结果。                //找出最大值        int maxNum=Integer.MIN_VALUE;        for(int n:nums){            if(n>maxNum)                maxNum=n;        }                //采用一行三列数组来记录，出现次数和首末位置下标        int [][] count=new int[maxNum+1][3];        int max=0;        for(int i=0;i<nums.length;i++){            if(count[nums[i]][0]==0){                //第一次出现，记录起始下标                count[nums[i]][1]=i;            }            //更新末尾下标            count[nums[i]][2]=i;                        count[nums[i]][0]++;            max=Math.max(max,count[nums[i]][0]);        }                //查找相同频率的元素        HashSet<Integer> set=new HashSet<Integer>();        //maxNum为了少判断一些数        for(int i=0;i<=maxNum;i++){            if(count[i][0]==max){                set.add(i);            }        }                //将最小间距的输出        int min=nums.length;        for(int index:set){            min=Math.min(min,count[index][2]-count[index][1]+1);        }                return min;    }}