LWC 54:697. Degree of an Array
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LWC 54:697. Degree of an Array
传送门:697. Degree of an Array
Problem:
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.
思路:
可以确定,答案一定是频次出现最多的元素,在数组中的最左边界和最右边界之差。当然频次出现最多的元素可能有多个,遍历一遍这些元素即可。
代码如下:
public int findShortestSubArray(int[] nums) { Map<Integer, Integer> lf = new HashMap<>(); Map<Integer, Integer> rt = new HashMap<>(); Map<Integer, Integer> cnt = new HashMap<>(); for (int i = 0; i < nums.length; ++i) { if (!lf.containsKey(nums[i])) lf.put(nums[i], i); rt.put(nums[i], i); cnt.put(nums[i], cnt.getOrDefault(nums[i], 0) + 1); } int degree = Collections.max(cnt.values()); int ans = Integer.MAX_VALUE; for (int key : cnt.keySet()) { if (degree == cnt.get(key)) { ans = Math.min(ans, rt.get(key) - lf.get(key) + 1); } } return ans; }
尺取法
构造合法窗口,在合法窗口下更新最小长度。时间复杂度O(n)
可参考博文:
http://blog.csdn.net/u014688145/article/details/73801021
代码如下:
public int findShortestSubArray(int[] nums) { Map<Integer, Integer> map = new HashMap<>(); int max = 0; for (int num : nums) { map.put(num, map.getOrDefault(num, 0) + 1); } for (int key : map.keySet()) { max = Math.max(max, map.get(key)); } int ans = 1 << 29; Map<Integer, Integer> cnt = new HashMap<>(); int lf = 0, rt = 0; int window = 0; for (;;) { while (rt < nums.length && window != max) { cnt.put(nums[rt], cnt.getOrDefault(nums[rt], 0) + 1); window = Math.max(window, cnt.get(nums[rt])); rt ++; } if (window != max) break; ans = Math.min(ans, rt - lf); cnt.put(nums[lf], cnt.get(nums[lf]) - 1); lf ++; window = Collections.max(cnt.values()); } return ans; }
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