LeetCode #697 - Degree of an Array

原创于 2019-02-05 22:33:11 发布 · 109 阅读

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本文介绍了一种高效算法，用于找到一个数组中度（即任意元素最大出现次数）与原数组相等的最短子数组。通过使用三个哈希映射，分别记录元素的计数、首次出现位置和最后出现位置，算法能在一次遍历中完成任务。

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.

一个数组的度的定义是频率最高元素出现的次数。求一个数组中最短的，度和原数组相同的子数组。为了求出数组的度，需要统计每个元素的次数，同时记录每个元素出现的第一个下标和最后一个下标，这样在遍历数组的过程中可以同时更新元素次数，一旦发现频率最高的元素，可以更新子数组的长度。

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map<int,int> count;
        unordered_map<int,int> first_index;
        unordered_map<int,int> last_index;
        int max_count=0;
        int min_length=INT_MAX;
        for(int i=0;i<nums.size();i++)
        {
            if(first_index.count(nums[i])==0) first_index[nums[i]]=i;
            last_index[nums[i]]=i;
            count[nums[i]]++;
            if(count[nums[i]]==max_count)
                min_length=min(min_length,last_index[nums[i]]-first_index[nums[i]]+1);
            else if(count[nums[i]]>max_count)
            {
                max_count=count[nums[i]];
                min_length=last_index[nums[i]]-first_index[nums[i]]+1;
            }
        }
        return min_length;
    }    
};