697. Degree of an Array*

697. Degree of an Array*

​​https://leetcode.com/problems/degree-of-an-array/​​

题目描述

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.

C++ 实现 1

来自 ​​[Java/C++/Python] One Pass Solution
​, 思路是:

在遍历 ​​A​​​ 的过程中, 统计每个元素出现的次数, 以及首次出现的 ​​index​​​, 之后不断更新 ​​degree​​ 以及 subarray 的 size. 用了两个哈希表.

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

C++ 实现 2

我的非常暴力的解法, 速度非常慢. 思路:

• 统计最大的 degree, 将拥有最大 degree 的元素保存到​​arr​​ 中
• 使用双指针找到最小的 subarray 的大小.

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map<int, int> records;
        unordered_set<int> arr;
        for (auto &a : nums) records[a] ++;
        int degree = 0;
        for (auto &p : records) degree = std::max(degree, p.second);
        for (auto &p : records) {
            if (p.second == degree) arr.insert(p.first);
        }
        int res = INT32_MAX;
        for (auto &p : arr) {
            int i = 0, j = nums.size() - 1;
            while (i <= j) {
                while (i <= j && nums[i] != p) ++ i;
                while (i <= j && nums[j] != p) -- j;
                res = std::min(res, j - i + 1);
                break;
            }
        }
        return res;
    }
};

C++ 实现 3

来自 LeetCode Submission. 和 ​​C++ 实现 1​​ 思路一样, 不过哈希表定义为:

unordered_map<int, pair<int, int>>

代码如下:

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        int ret = INT_MAX, cur_max = 0;
        unordered_map<int, pair<int, int>> m;
        for(int i = 0; i < nums.size(); i++){
            if(m.find(nums[i]) == m.end()){
                m[nums[i]] = make_pair(0, i);
            }
            m[nums[i]].first++;
            int degree = m[nums[i]].first, dist = i - m[nums[i]].second + 1;
            if(cur_max < degree){
                cur_max = degree;
                ret = dist;
            }
            else if(cur_max == degree){
                ret = min(ret, dist);
            }
        }
        return ret;
    }
};