leetCode-Degree of an Array

Description:
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.

My Solution:

class Solution {    public int findShortestSubArray(int[] nums) {        Map<Integer,Integer>map = new HashMap<Integer,Integer>();        int len = nums.length;        for(int i = 0;i < len;i++){            map.put(nums[i],(map.get(nums[i]) == null)?0:map.get(nums[i])+1);        }        int maxCount = 0;        List<Integer>list = new ArrayList<Integer>();        for(Integer key : map.keySet()){            if(map.get(key) > maxCount){                maxCount = map.get(key);            }        }        for(Integer key : map.keySet()){            if(map.get(key) == maxCount){                list.add(key);            }        }        List<Integer> temp = new ArrayList<Integer>();        for(int i:nums){            temp.add(Integer.valueOf(i));        }        int shortest = temp.lastIndexOf(list.get(0)) - temp.indexOf(list.get(0)) + 1;        for(int i = 0;i < list.size();i++){            int first = temp.indexOf(list.get(i));            int last = temp.lastIndexOf(list.get(i));            int length = last - first +1;            if(length < shortest){                shortest = length;            }        }        return shortest;    }}

Better Solution:

class Solution {    public int findShortestSubArray(int[] nums) {           if (nums.length == 0 || nums == null) return 0;        Map<Integer, int[]> map = new HashMap<>();        for (int i = 0; i < nums.length; i++){           if (!map.containsKey(nums[i])){               map.put(nums[i], new int[]{1, i, i});  // the first element in array is degree, second is first index of this key, third is last index of this key           } else {               int[] temp = map.get(nums[i]);               temp[0]++;               temp[2] = i;           }        }        int degree = Integer.MIN_VALUE, res = Integer.MAX_VALUE;        for (int[] value : map.values()){            if (value[0] > degree){                degree = value[0];                res = value[2] - value[1] + 1;            } else if (value[0] == degree){                res = Math.min( value[2] - value[1] + 1, res);            }         }        return res;    }}

Best Solution:

class Solution {    public int findShortestSubArray(int[] nums) {        if (nums == null || nums.length == 0)            return 0;        int maxNum = 0;        for (int n : nums){            maxNum = Math.max(n, maxNum);        }        //由于数组的元素都为正整数,因此int[maxNum + 1]的下标可以包含数组中所有元素        //start以nums的元素为下标，保存对应的起始点        int[] start = new int[maxNum + 1];        //end以nums的元素为下标，保存对应的终止点        int[] end = new int[maxNum + 1];        //que以nums的元素为下标，保存对应的次数        int[] que = new int[maxNum + 1];        for (int i = 0; i < nums.length; i++){        //第一次出现nums[i]时，将i存入start,即起始点            if (que[nums[i]] == 0)                start[nums[i]] = i;            //每次出现同一个nums元素，更新end数组,即终止点            end[nums[i]] = i;            //每次出现同一个nums元素，更新que数组，计数            que[nums[i]]++;        }        int max = 0;        //找出最大degree        for (int n : que)            max = Math.max(max, n);        List<Integer> maxNums = new ArrayList<>();        #找出que中拥有最大计数的下标        for (int i = 0; i < que.length; i++){            if (que[i] == max)                maxNums.add(i);        }        int res = nums.length;        #遍历maxNums,找出满足degree的最小子集长度        for (int n : maxNums){            int r = end[n] - start[n] + 1;            res = Math.min(r, res);        }        return res;    }}