[leetcode] 697. Degree of an Array

Description

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.

分析

题目的意思是：给你一个数组，给了一个degree的定义：这个数组中元素的频率。求出与原数组中degree相同的子数组的最短长度。

• 用hash表存放次数，用另一个hash表存放该数值在数组的起始位置和结束位置。最后把degree的最大的几个数取出来，取最小长度就行了。

代码

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

参考文献

[LeetCode] Degree of an Array 数组的度