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.
Solution
HashMap, time O(n), space O(n)
一个loop即可，一边遍历一边更新ssaLength。
class Solution {
public int findShortestSubArray(int[] nums) {
int maxDegree = 0;
int ssaLength = 0;
Map<Integer, List<Integer>> numToIndexes = new HashMap<>();
for (int i = 0; i < nums.length; ++i) {
if (!numToIndexes.containsKey(nums[i])) {
numToIndexes.put(nums[i], new ArrayList<>());
}
List<Integer> indexList = numToIndexes.get(nums[i]);
indexList.add(i);
int length = indexList.get(indexList.size() - 1) - indexList.get(0) + 1;
if (indexList.size() > maxDegree
|| (indexList.size() == maxDegree && length < ssaLength)) {
maxDegree = indexList.size();
ssaLength = length;
}
}
return ssaLength;
}
}
