[LeetCode]Degree of an Array 数组的度

链接https://leetcode.com/problems/degree-of-an-array/description/

难度:Easy

题目:697. Degree of an Array

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:

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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:

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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.

翻译:给定一个非空非负的整型数组,定义数组的度为数组中元素出现的最大次数。任务是找出度和数组的度相同的最小子串

思路:记录下第一次出现和最后一次出现的位置就好了,两者相减就是最短长度。对于有多个出现次数最多元素的情况,只需要找出这些元素的最短子串中最小的就好了。

参考代码

Java

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class Solution {
public int findShortestSubArray(int[] nums) {
Map<Integer, Integer> left = new HashMap();
Map<Integer, Integer> right = new HashMap();
Map<Integer, Integer> count = new HashMap();

for(int i=0; i < nums.length; i++ ){
if(!left.containsKey(nums[i]))
left.put(nums[i], i);
right.put(nums[i], i);
count.put(nums[i], count.getOrDefault(nums[i],0)+1);
}

int degree = Collections.max(count.values());
int length = Integer.MAX_VALUE;
for(int i=0; i<nums.length; i++){
if(count.get(nums[i])==degree){
length = Math.min(length, right.get(nums[i]) - left.get(nums[i]) + 1);
}
}
return length;
}
}