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滑动窗口(三)——Longest K unique charac

滑动窗口(三)——Longest K unique charac

作者: GavinLaw | 来源:发表于2020-03-24 00:24 被阅读0次

    题目:

    Given a string you need to print the size of the longest possible substring that has exactly k unique characters. If there is no possible substring print -1.
    Example
    For the string aabacbebebe and k = 3 the substring will be cbebebe with length 7.

    Input:
    The first line of input contains an integer T denoting the no of test cases then T test cases follow. Each test case contains two lines . The first line of each test case contains a string s and the next line conatains an integer k.

    Output:
    For each test case in a new line print the required output.

    Constraints:
    1<=T<=100
    1<=k<=10

    Example:
    Input:
    2
    aabacbebebe
    3
    aaaa
    1
    Output:
    7
    4

    代码:

    import java.io.BufferedReader;
    import java.io.IOException;
    import java.io.InputStreamReader;
    
    public class GFG {
        public static void main(String args[]) throws IOException {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            int t = Integer.parseInt(br.readLine());
            while(t-->0){
                String str = br.readLine();
                int k = Integer.parseInt(br.readLine());
                int len = str.length();
                int [] arr = new int[26];
                int dis=0;
                int start=0;
                int maxLen=-1;
                for(int i=0;i<len;i++){
                    if(arr[str.charAt(i)-'a']==0){
                        dis++;
                    }
                    arr[str.charAt(i)-'a']++;
                    while(dis>k){
                        arr[str.charAt(start)-'a']--;
                        if(arr[str.charAt(start)-'a']==0){
                            dis--;
                        }
                        start++;
                    }
                    if(dis==k&&maxLen<i-start+1){
                        maxLen=i-start+1;
                    }
                }
                System.out.println(maxLen);
            }
        }
    }
    

    分析:

    这道题和滑动窗口(二)有点像,但比它稍微复杂点,首先也是用个arr存同种类的字符,然后滑动窗口,不同的是,这里如果匹配不到k个不同字符串,则输出-1,所以处理与前一篇略有不同.

    时间复杂度同样是O(N)

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