697. Degree of an Array

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

方法一：
用hash table 统计所有number出现的次数，找出最高频率以及所有出现最高频率的数字，并依次计算他们在原数组中的左右边界，也就是第一次和最后一个出现的位置。

class Solution {public:    int findShortestSubArray(vector<int>& nums) {        unordered_map<int, int> hash;        int maxfreq = 0;        for (int i = 0; i < nums.size(); i++) {            hash[nums[i]]++;            maxfreq = max(hash[nums[i]], maxfreq);        }        if (maxfreq == 1) return 1;        vector<int> max;        for (auto m : hash) {            if (m.second == maxfreq) {                max.push_back(m.first);            }        }        int minlen = nums.size();        for (int i = 0; i < max.size(); i++) {            int cand = max[i];            int minId = -1;            int maxId = -1;            int freq = maxfreq;            for (int j = 0; j < nums.size() && freq > 0; j++) {                if (nums[j] == cand) {                    freq--;                    if (minId == -1) minId = j;                    if (freq == 0) maxId = j;                }            }            if (maxId - minId + 1 < minlen) {                minlen = maxId - minId + 1;            }        }        return minlen;    }};

方法二：
用空间换时间，减少遍历的次数：
多建立两个hash table 存储每一个number的左右边界下标，遍历一遍数组，建立三个hash table.

class Solution {public:    int findShortestSubArray(vector<int>& nums) {        unordered_map<int, int> hash_cnt;        unordered_map<int, int> hash_left;        unordered_map<int, int> hash_right;        int maxfreq = 0;        for (int i = 0; i < nums.size(); i++) {            hash_cnt[nums[i]]++;            maxfreq = max(hash_cnt[nums[i]], maxfreq);            if (hash_left.find(nums[i]) == hash_left.end()) {                hash_left[nums[i]] = i;            }            hash_right[nums[i]] = i;        }        if (maxfreq == 1) return 1;        int minlen = nums.size();        for (auto m : hash_cnt) {            if (m.second == maxfreq) {                minlen = min(minlen, hash_right[m.first] - hash_left[m.first] + 1);            }        }        return minlen;    }};