递归与分治策略-2.6棋盘覆盖

public class test2_6 {
    static int tile = 0;
    static int size = 4;
    static int[][] board = new int[size][size];
    public static void chessBoard(int H,int L,int TH,int TL,int size){
        if(size==1) return;
        int t = tile++;
        int s = size/2;
        //1.左上角
        //特殊方格在此棋盘里
        if(TH<H+s&&TL<L+s){
            chessBoard(H,L,TH,TL,s);
        }else{//特殊方格不在此棋盘里
            board[H+s-1][L+s-1] = t;
            chessBoard(H,L,H+s-1,L+s-1,s);
        }
        //2.右上角
        //特殊方格在此棋盘里
        if(TH<H+s&&TL>=L+s){
            chessBoard(H,L+s,TH,TL,s);
        }else{//特殊方格不在此棋盘里
            board[H+s-1][L+s] = t;
            chessBoard(H,L+s,H+s-1,L+s,s);
        }
        //3.左下角
        //特殊方格在此棋盘里
        if(TH>=H+s&&TL<L+s){
            chessBoard(H+s,L,TH,TL,s);
        }else{//特殊方格不在此棋盘里
            board[H+s][L+s-1] = t;
            chessBoard(H+s,L,H+s,L+s-1,s);
        }   
        //4.右下角
        //特殊方格在此棋盘里
        if(TH>=H+s&&TL>=L+s){
            chessBoard(H+s,L+s,TH,TL,s);
        }else{//特殊方格不在此棋盘里
            board[H+s][L+s] = t;
            chessBoard(H+s,L+s,H+s,L+s,s);
        }   
    }
    public static void main(String[] args) {
        int H=0,L=0,TH=0,TL=1;
        chessBoard(H,L,TH,TL,size);
        for(int i=0;i<size;i++){
            for(int j=0;j<size;j++)
                System.out.print(board[i][j]+" ");
            System.out.println();
        }
    }
}

输出结果如下:

1 0 2 2 
1 1 0 2 
3 0 0 4 
3 3 4 4 
    原文作者:递归与分治算法
    原文地址: https://blog.csdn.net/SL_World/article/details/78171466
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