[LeetCode] Degree of an Array

Posted 2020-10-11 immjc

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.

题目要求给定一个非空非负的整数数组，先求出数组的度，也就是出现次数最多的那个数字的出现次数。

然后找出这个数组的最小连续子数组并返回子数组的长度，要求这个子数组的度和原数组的度相同。

这是个很容易的题。就写了一个思路简单但是不够简洁的方法。

思路：首先利用map(m)求出给定数组的度，这里需要注意的是有可能含有相同度的多个数字。所以用一个数组vec来存储数组的度相同的数字。然后利用另一个map(n)来存储数组中从尾到头不同数字的索引（遍历数组，用 map映射数值到索引，重复的数字就会被覆盖并更新索引），这个索引也就是子数组的右边界。最后利用两层for循环，找出子数组的左边界。并返回最小的子数组长度即可。

想法比较片段化，后续想出更简单的方法再更新。

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        int degree = 0, val = 0, res = INT_MAX;
        unordered_map<int, int> m;
        for (int num : nums) 
            m[num]++;
        for (auto it = m.begin(); it != m.end(); it++) {
            if (it->second > degree) {
                degree = it->second;
            }
        }
        vector<int> vec;
        for (auto it = m.begin(); it != m.end(); it++) {
            if (it->second == degree) {
                vec.push_back(it->first);
            }
        }
        unordered_map<int, int> n;
        for (int i = 0; i != nums.size(); i++) 
            n[nums[i]] = i;
        for (int i = 0; i != vec.size(); i++) {
            int left = 0, right = n[vec[i]];
            bool first = true;
            for (int j = 0; j != nums.size(); j++) {
                if (vec[i] == nums[j] && first) {
                    left = j;
                    first = false;
                }
                res = min(res, right - left + 1);
            }
        }
        return res;
    }
};
// 279 ms