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¶
数组的度。最直接的方法:利用哈希表找到数组的度,然后找到度的对应数字,寻找这些数字的长度的最小值。
public int findShortestSubArray(int[] nums) { if (nums == null || nums.length == 0) return 0; // map中的键是nums中的数字,值是在nums中出现的位置 Map<Integer, List<Integer>> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { int num = nums[i]; if (!map.containsKey(num)) map.put(num, new ArrayList<>()); map.get(num).add(i); } // 计算数组nums的degree int degree = 0; for (int num : map.keySet()) degree = Math.max(degree, map.get(num).size()); // 计算在nums中达到最大degree的数字 List<Integer> degreeList = new ArrayList<>(); for (int num : map.keySet()) if (degree == map.get(num).size()) degreeList.add(num); // 计算最小长度 int minLength = nums.length; for (int num : degreeList) { List<Integer> positions = map.get(num); Collections.sort(positions); int length = positions.get(positions.size() - 1) - positions.get(0) + 1; if (length < minLength) minLength = length; } return minLength; }
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