题解 | #DNA序列#

使用滑动窗口解决，维护一个长度大小固定为N的窗口

import java.util.*;

public class Main{
    
    public static void main(String[] args){
        
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            String DNA = sc.nextLine();
            int N = sc.nextInt();
            String gc = maxGC(DNA, N);
            System.out.println(gc);
        }
    }
    //使用滑动串口解决，滑动窗口大小固定为N
    public static String maxGC(String DNA, int N){
        if(N >= DNA.length()){
            return DNA;
        }
        int g = 0, c = 0;//分别表示G和C的数目
        //初始化滑动窗口
        int left = 0, right = 0;
        double maxGCRate = 0.0;
        String gc = "";
        while(right < N){
            if(DNA.charAt(right) == 'G'){
                g++;
            }else if(DNA.charAt(right) == 'C'){
                c++;
            }
            right++;
        }
       
        while(right < DNA.length()){
            double gcRate = (g + c) * 1.0 / N;
            if(gcRate > maxGCRate){
                maxGCRate = gcRate;
                gc = DNA.substring(left, left + N);
            }
            if(DNA.charAt(left) == 'G'){
                g--;
            }else if(DNA.charAt(left) == 'C'){
                c--;
            }
            left++;
            if(DNA.charAt(right) == 'G'){
                g++;
            }else if(DNA.charAt(right) == 'C'){
                c++;
            }
            right++;
        }
        return gc;
    }
}