697. Degree of an Array

标签：++   vector   count   and   order   思路   one   tween   short   

Problem:

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.

思路：

Solution (C++):

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

性能：

Runtime: 60 ms??Memory Usage: 10.9 MB

思路：

Solution (C++):

性能：

Runtime: ms??Memory Usage: MB

697. Degree of an Array

标签：++   vector   count   and   order   思路   one   tween   short   

原文地址：https://www.cnblogs.com/dysjtu1995/p/12727137.html