697. Degree of an Array

问题: 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.

问题分析： 输入一个字符串，输出具有最多相同字符的子串的最短长度。

解题思路： 用哈希映射数据结构来解决问题，在键中存储数组元素，重新定义一个包含起始位置begin、长度length、与相同个数times的数据类型，用于值中。最后对值进行选取，先判断个数最多的值，其次再次数最多且相等的情况下，选择长度最短长度，即是所要的输出。

*class Solution {                    public int findShortestSubArray(int[] nums) {                         Map<Integer,Beginlength> map = new HashMap<Integer,Beginlength>();                        int begin ;                        int length = 0;                        int times = 0;                        for(int i = 0; i < nums.length;i++){                                     if(!map.containsKey(nums[i])){                                begin = i;                                length = 1;                                times = 1;                            }                            else{                                begin = map.get(nums[i]).begin;                                times = ++ map.get(nums[i]).times;                                length = i - begin + 1;                            }                               map.put(nums[i], new Beginlength(begin,length,times)); //传递实例化对象                        }                        int j = 0;                        int time = 0;                        for(Beginlength i:map.values()){                        if(i.times > time){                              time = i.times;                              j = i.length;                            }                        else if(i.times == time){                            j =(j>i.length)?i.length:j;                          }                        }                        return j;                    }}        class Beginlength{            int begin ;            int length;            int times;            Beginlength(int begin, int length,int times){                this.begin = begin;                this.length = length;                this.times = times;            }    }   

运行时间为49ms。