[LeetCode] 697. Degree of an Array 数组的度

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.

给一个非空非负整数的数组，找出和整个数组最大度相同的最短的子数组。度是一个数组里数字出现频率的最大值。

解法：首先要知道整个数组的度，可用哈希表进行统计。度最大的子数组里，首尾数字都是这个度的数字时，子数组是最短的，在用一个哈希表保存，某一个数字第一次出现的index和最后一次出现的index。还要注意度相同的数字可能不只一个所以要求这几个数字中最短的。

Java:

public static class Solution2 {
    public int findShortestSubArray(int[] nums) {
        Map<Integer, Integer> count = new HashMap<>();
        Map<Integer, Integer> left = new HashMap<>();
        Map<Integer, Integer> right = new HashMap<>();

        for (int i = 0; i < nums.length; i++) {
            count.put(nums[i], count.getOrDefault(nums[i], 0) + 1);
            if (!left.containsKey(nums[i])) {
                left.put(nums[i], i);
            }
            right.put(nums[i], i);
        }

        int result = nums.length;
        int degree = Collections.max(count.values());
        for (int num : count.keySet()) {
            if (count.get(num) == degree) {
                result = Math.min(result, right.get(num) - left.get(num) + 1);
            }
        }
        return result;
    }
}

Java:

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;
    }　　

Python:

class Solution(object):
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        counts = collections.Counter(nums)
        left, right = {}, {}
        for i, num in enumerate(nums):
            left.setdefault(num, i)
            right[num] = i
        degree = max(counts.values())
        return min(right[num]-left[num]+1 \
                   for num in counts.keys() \
                   if counts[num] == degree)

Python: wo

class Solution(object):
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) < 2:
            return len(nums)

        degree = collections.Counter()
        index = collections.defaultdict(list)
        max_degree = 0
        max_num = []
        for k, v in enumerate(nums):
            degree[v] += 1
            if degree[v] > max_degree:
                max_degree = degree[v]
                max_num = []
                max_num.append(v)
            elif degree[v] == max_degree:
                max_num.append(v)
            index[v].append(k)

        min_dis = float('inf')
        for v in max_num:
            i, j = index[v][0], index[v][-1]
            min_dis = min(min_dis, j - i + 1)

        return min_dis

C++:

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        int n = nums.size(), res = INT_MAX, degree = 0;
        unordered_map<int, int> m;
        unordered_map<int, pair<int, int>> pos;
        for (int i = 0; i < nums.size(); ++i) {
            if (++m[nums[i]] == 1) {
                pos[nums[i]] = {i, i};
            } else {
                pos[nums[i]].second = i;
            }
            degree = max(degree, m[nums[i]]);
        }
        for (auto a : m) {
            if (degree == a.second) {
                res = min(res, pos[a.first].second - pos[a.first].first + 1);
            }
        }
        return res;
    }
};

C++:

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        int n = nums.size(), res = INT_MAX, degree = 0;
        unordered_map<int, int> m, startIdx;
        for (int i = 0; i < n; ++i) {
            ++m[nums[i]];
            if (!startIdx.count(nums[i])) startIdx[nums[i]] = i;
            if (m[nums[i]] == degree) {
                res = min(res, i - startIdx[nums[i]] + 1);
            } else if (m[nums[i]] > degree) {
                res = i - startIdx[nums[i]] + 1;
                degree = m[nums[i]];
            }
        }
        return res;
    }
};

　　

　　

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