LeetCode-697：Degree of an Array (度相同的最小子数组)

Question

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.

问题解析：

给定非空非负整数数组，数组的度是指数组中出现次数最多元素的个数。寻找最小连续子数组，使得子数组的度与原数组的度相同。返回子数组的长度。

Answer

Solution 1：

利用数据结构实现，Map、Dict。

• maxs记录元素的最大下标，mins记录元素的最小下标，cnts记录元素的出现个数，三者均是HashMap结构；
• 通过遍历记录元素与上三个对应的记录，只要找到最大次数对应的元素，则就找到了该元素的最大和最小下标，即找到了度相同的最小子数组。

class Solution {
    public int findShortestSubArray(int[] nums) {

        Map<Integer, Integer> start = new HashMap<Integer, Integer>();
        Map<Integer, Integer> end = new HashMap<Integer, Integer>();
        Map<Integer, Integer> count  = new HashMap<Integer, Integer>();

        int maxCount = 0;
        for (int i = 0; i < nums.length; i++){
            if (!count.containsKey(nums[i])){
                count.put(nums[i], 0);
                start.put(nums[i], i);
            }
            count.put(nums[i], count.get(nums[i])+1);
            end.put(nums[i], i);
            maxCount = Math.max(maxCount, count.get(nums[i]));
        }

        int minLength = Integer.MAX_VALUE;
        for (Integer key : count.keySet()){
            if (count.get(key) == maxCount){
                minLength = Math.min(minLength,end.get(key)-start.get(key)+1);
            }
        }

        return minLength;
    }
}

• Runtime: 82 ms
• Beats 28.81 % of java submissions
• 时间复杂度：O(n)；空间复杂度：O(1)