LWC 54：697. Degree of an Array

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：
javascript:void(0)

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;
    }