友情支持

如果您觉得这个笔记对您有所帮助,看在D瓜哥码这么多字的辛苦上,请友情支持一下,D瓜哥感激不尽,😜

支付宝

微信

有些打赏的朋友希望可以加个好友,欢迎关注D 瓜哥的微信公众号,这样就可以通过公众号的回复直接给我发信息。

wx jikerizhi

公众号的微信号是: jikerizhi因为众所周知的原因,有时图片加载不出来。 如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。

992. Subarrays with K Different Integers

Given an array A of positive integers, call a (contiguous, not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly K.

(For example, [1,2,3,1,2] has 3 different integers: 1, 2, and 3.)

Return the number of good subarrays of A.

Example 1:
Input: A = [1,2,1,2,3], K = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].
Example 2:
Input: A = [1,2,1,3,4], K = 3
Output: 3
Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].

Note:

  • 1 ⇐ A.length ⇐ 20000

  • 1 ⇐ A[i] ⇐ A.length

  • 1 ⇐ K ⇐ A.length

解题分析

解题中心思路是滑动窗口。

思考题

这个解题方案怎么理解?

参考资料

Given an array A of positive integers, call a (contiguous, not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly K.

(For example, [1,2,3,1,2] has 3 different integers: 1, 2, and 3.)

Return the number of good subarrays of A.

Example 1:

Input: A = [1,2,1,2,3], K = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].

Example 2:

Input: A = [1,2,1,3,4], K = 3
Output: 3
Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].

Note:

  1. 1 ⇐ A.length ⇐ 20000

  2. 1 ⇐ A[i] ⇐ A.length

  3. 1 ⇐ K ⇐ A.length

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
/**
 * Runtime: 5 ms, faster than 93.74% of Java online submissions for Subarrays with K Different Integers.
 * Memory Usage: 51.9 MB, less than 5.26% of Java online submissions for Subarrays with K Different Integers.
 *
 * Copy from: https://mp.weixin.qq.com/s/6YeZUCYj5ft-OGa85sQegw[面试官,你再问我滑动窗口问题试试?我有解题模板,不怕!]
 *
 * @author D瓜哥 · https://www.diguage.com
 * @since 2020-01-30 19:44
 */
public int subarraysWithKDistinct(int[] A, int K) {
    if (Objects.isNull(A) || A.length == 0 || A.length < K) {
        return 0;
    }
    int[] hash = new int[A.length + 1];
    int left = 0, count = 0;
    int results = 0, result = 1;
    for (int right = 0; right < A.length; right++) {
        hash[A[right]]++;
        if (hash[A[right]] == 1) {
            count++;
        }
        while (hash[A[left]] > 1 || count > K) {
            if (count > K) {
                result = 1;
                count--;
            } else {
                result++;
            }
            hash[A[left]]--;
            left++;
        }
        if (count == K) {
            results += result;
        }
    }
    return results;
}

/**
 * Runtime: 53 ms, faster than 67.75% of Java online submissions for Subarrays with K Different Integers.
 * Memory Usage: 43.9 MB, less than 10.53% of Java online submissions for Subarrays with K Different Integers.
 *
 * Copy from: https://leetcode-cn.com/problems/subarrays-with-k-different-integers/solution/k-ge-bu-tong-zheng-shu-de-zi-shu-zu-by-leetcode/[K 个不同整数的子数组 - K 个不同整数的子数组 - 力扣(LeetCode)]
 */
public int subarraysWithKDistinctWindows(int[] A, int K) {
    if (Objects.isNull(A) || A.length == 0 || A.length < K) {
        return 0;
    }
    Window w1 = new Window();
    Window w2 = new Window();
    int result = 0;
    int left1 = 0, left2 = 0;
    for (int right = 0; right < A.length; right++) {
        int x = A[right];
        w1.add(x);
        w2.add(x);
        while (w1.different() > K) {
            w1.remove(A[left1++]);
        }
        while (w2.different() >= K) {
            w2.remove(A[left2++]);
        }
        result += left2 - left1;
    }
    return result;
}

private class Window {
    Map<Integer, Integer> count;
    int nonzeo;

    public Window() {
        count = new HashMap<>();
        nonzeo = 0;
    }

    void add(int x) {
        count.put(x, count.getOrDefault(x, 0) + 1);
        if (count.get(x) == 1) {
            nonzeo++;
        }
    }

    void remove(int x) {
        count.put(x, count.get(x) - 1);
        if (count.get(x) == 0) {
            nonzeo--;
        }
    }

    int different() {
        return nonzeo;
    }
}