[算法]循环打印矩阵，蛇形矩阵专题

Posted 2020-10-30 陈驰字新宇

1. 循环打印矩阵

比方提供以下矩阵：

按照如下顺序打印出来：

1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10

这道题直接写也没问题，就是特别容易出错，稍不留意就写错，而且这类题型我想要一种普适性的解法。

我想到的一种方法就是一圈一圈打印，从外到内，我们确定一个矩形，通常通过左上角的坐标和右下角的坐标即可，即（tR，tC）和（dR，dC），我们先写出打印一圈的方法，然后循环调用，如果我们发现左上角的坐标跑到了右下角坐标的右边或者下边，整个过程就停止，这样额外的空间复杂度是O(1)。No bibi，show me the code.

package com.darrenchan;

/**
 * 转圈打印矩阵
 * @author Think
 *
 */
public class SheXingPrint2 {
    public static void main(String[] args) {
        int[][] m = new int[][]{{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}};
        
        int tR = 0, tC = 0, dR = 0, dC = 3;
        while(tR <= dR && tC <= dC){
            printEdge(m, tR++, tC++, dR--, dC--);
        }
    }
    
    public static void printEdge(int[][] m, int tR, int tC, int dR, int dC){
        if(tR == dR){
            for (int i = tC; i <= dC; i++) {
                System.out.print(m[tR][i] + " ");
            }
        }else if(tC == dC){
            for (int i = tR; i <= dR; i++) {
                System.out.print(m[i][tC] + " ");
            }
        }else{
            int curRow = tR;
            int curCol = tC;
            while(curCol != dC){
                System.out.print(m[tR][curCol] + " ");
                curCol++;
            }
            while(curRow != dR){
                System.out.print(m[curRow][dC] + " ");
                curRow++;
            }
            
            while(curCol != tC){
                System.out.print(m[dR][curCol] + " ");
                curCol--;
            }
            
            while(curRow != tR){
                System.out.print(m[curRow][tC] + " ");
                curRow--;
            }
            
        }
    }
}

2. 蛇形矩阵

给出一个数字，比方是4，直接生成如下矩阵：

思路和上一个类似，只是稍微改一下即可：

package com.darrenchan;

/**
 * 打印蛇形矩阵
 * @author Think
 *
 */
public class SheXingPrint3 {
    public static int num = 1;
    public static void main(String[] args) {
        //给出多大的数字，就构造多大的矩阵
        int length = 4;
        int[][] m = new int[length][length];

        int tR = 0, tC = 0, dR = m.length - 1, dC = m[0].length - 1;
        while (tR <= dR && tC <= dC) {
            generateEdge(m, tR++, tC++, dR--, dC--);
        }
        
        for (int i = 0; i < m.length; i++) {
            for (int j = 0; j < m[i].length; j++) {
                System.out.print(m[i][j] + "\\t");
            }
            System.out.println();
        }
    }

    public static void generateEdge(int[][] m, int tR, int tC, int dR, int dC) {
        if(tR == dR){//处理length为奇数，最中间一个元素
            m[tR][tC] = num;
        }else{
            int curRow = tR;
            int curCol = tC;
            while(curCol != dC){
                m[tR][curCol] = num;
                num++;
                curCol++;
            }
            while(curRow != dR){
                m[curRow][dC] = num;
                num++;
                curRow++;
            }
            while(curCol != tC){
                m[dR][curCol] = num;
                num++;
                curCol--;
            }
            while(curRow != tR){
                m[curRow][tC] = num;
                num++;
                curRow--;
            }
        }
    }
}

 

3. 双循环打印矩阵

一个起始点在左上角，和以前一样，顺时针走，另一个起始点在右下角，也是顺时针走，别人走过的，就不能重复走了。

比方

打印出：1 2 3 4 

打印出：1 2 3 6 5 4 7 8 9 

打印出：1 2 3 4 8 12 11 10 7 6 5 9 13 14 15 16 

这道题的思路其实和上面第一个差不多，我还是采用的一圈一圈打印的思路，只不过原来一个指针打印一圈转4个方向，现在有两个指针，每个转2个方向，第一次转两个方向，第二次转另两个方向，依次类推。

其实这道题还可以进一步优化，因为两个指针完全关于矩阵中心点对称，所以确定一个，另一个就确定了。这里需要一个矩阵来记录该点是否已经走过。

代码如下：

package com.darrenchan;

import java.util.LinkedList;
import java.util.List;

/**
 * 双蛇形打印
 * @author Think
 *
 */
public class SheXingPrint4 {
    public static int num = 1;
    public static void main(String[] args) {
        //给出多大的数字，就构造多大的矩阵
        int length = 5;
        int[][] m = new int[length][length];
        int temp = 1;
        for (int i = 0; i < m.length; i++) {
            for (int j = 0; j < m[i].length; j++) {
                m[i][j] = temp++;
            }
        }

        int[][] visited = new int[m.length][m[0].length];//0是没有访问过，1是访问过
        
        List<Integer> list1 = new LinkedList<>();
        List<Integer> list2 = new LinkedList<>();        
        
        int tr = 0;
        int tc = 0;
        int dr = m.length - 1;
        int dc = m[0].length - 1;
        
        int count = 1;
        while(tr <= dr && tc <= dc){
            
            if(count % 2 == 1){
                int curRow1 = tr;
                int curCol1 = tc;
                int curRow2 = dr;
                int curCol2 = dc;
                while(curCol1 != dc && curCol2 != tc){
                    if(visited[curRow1][curCol1] == 0 && visited[curRow2][curCol2] == 0){
                        visited[curRow1][curCol1] = 1;
                        list1.add(m[curRow1][curCol1]);
                        curCol1++;
                        visited[curRow2][curCol2] = 1;
                        list2.add(0, m[curRow2][curCol2]);
                        curCol2--;
                    }else{
                        break;
                    }
                }
                
                while(curRow1 != dr && curRow2 != tr){
                    if(visited[curRow1][curCol1] == 0 && visited[curRow2][curCol2] == 0){
                        visited[curRow1][curCol1] = 1;
                        list1.add(m[curRow1][curCol1]);
                        curRow1++;
                        visited[curRow2][curCol2] = 1;
                        list2.add(0, m[curRow2][curCol2]);
                        curRow2--;
                    }else{
                        break;
                    }
                }
            }
            
            if(count % 2 == 0){
                int curRow1 = dr;
                int curCol1 = dc;
                int curRow2 = tr;
                int curCol2 = tc;
                while(curCol1 != tc && curCol2 != dc){
                    if(visited[curRow1][curCol1] == 0 && visited[curRow2][curCol2] == 0){
                        visited[curRow1][curCol1] = 1;
                        list1.add(m[curRow1][curCol1]);
                        curCol1--;
                        visited[curRow2][curCol2] = 1;
                        list2.add(0, m[curRow2][curCol2]);
                        curCol2++;
                    }else{
                        break;
                    }
                }
                
                while(curRow1 != tr && curRow2 != dr){
                    if(visited[curRow1][curCol1] == 0 && visited[curRow2][curCol2] == 0){
                        visited[curRow1][curCol1] = 1;
                        list1.add(m[curRow1][curCol1]);
                        curRow1--;
                        visited[curRow2][curCol2] = 1;
                        list2.add(0, m[curRow2][curCol2]);
                        curRow2++;
                    }else{
                        break;
                    }
                }
            }
            
            count++;
            tr++;
            tc++;
            dr--;
            dc--;
        }
        
        System.out.println(list1);
        System.out.println(list2);
        if(length % 2 == 0){
            for (int i = 0; i < list1.size(); i++) {
                System.out.print(list1.get(i) + " ");
            }
            for (int i = 0; i < list2.size(); i++) {
                System.out.print(list2.get(i) + " ");
            }
        }else{
            for (int i = 0; i < list1.size(); i++) {
                System.out.print(list1.get(i) + " ");
            }
            System.out.print(m[(length - 1) / 2][(length - 1) / 2] + " ");
            for (int i = 0; i < list2.size(); i++) {
                System.out.print(list2.get(i) + " ");
            }
        }
    }

}

优化之后的代码如下：

package com.darrenchan;

import java.util.LinkedList;
import java.util.List;

/**
 * 双蛇形打印(优化版)
 * @author Think
 *
 */

public class SheXingPrint5 {
    public static int num = 1;
    public static void main(String[] args) {
        //给出多大的数字，就构造多大的矩阵
        int length = 4;
        int[][] m = new int[length][length];
        int temp = 1;
        for (int i = 0; i < m.length; i++) {
            for (int j = 0; j < m[i].length; j++) {
                m[i][j] = temp++;
            }
        }

        int[][] visited = new int[m.length][m[0].length];//0是没有访问过，1是访问过
        
        List<Integer> list1 = new LinkedList<>();
        List<Integer> list2 = new LinkedList<>();        
        
        int tr = 0;
        int tc = 0;
        int dr = m.length - 1;
        int dc = m[0].length - 1;
        
        int count = 1;
        while(tr <= dr && tc <= dc){
            
            if(count % 2 == 1){
                int curRow1 = tr;
                int curCol1 = tc;
                while(curCol1 != dc){
                    if(visited[curRow1][curCol1] == 0){
                        visited[curRow1][curCol1] = 1;
                        list1.add(m[curRow1][curCol1]);
                        visited[length - 1 - curRow1][length - 1 - curCol1] = 1;
                        list2.add(0, m[length - 1 - curRow1][length - 1 - curCol1]);
                        curCol1++;
                    }else{
                        break;
                    }
                }
                
                while(curRow1 != dr){
                    if(visited[curRow1][curCol1] == 0){
                        visited[curRow1][curCol1] = 1;
                        list1.add(m[curRow1][curCol1]);
                        visited[length - 1 - curRow1][length - 1 - curCol1] = 1;
                        list2.add(0, m[length - 1 - curRow1][length - 1 - curCol1]);
                        curRow1++;
                    }else{
                        break;
                    }
                }
            }
            
            if(count % 2 == 0){
                int curRow1 = dr;
                int curCol1 = dc;
                while(curCol1 != tc){
                    if(visited[curRow1][curCol1] == 0){
                        visited[curRow1][curCol1] = 1;
                        list1.add(m[curRow1][curCol1]);
                        visited[length - 1 - curRow1][length - 1 - curCol1] = 1;
                        list2.add(0, m[length - 1 - curRow1][length - 1 - curCol1]);
                        curCol1--;
                    }else{
                        break;
                    }
                }
                
                while(curRow1 != tr){
                    if(visited[curRow1][curCol1] == 0){
                        visited[curRow1][curCol1] = 1;
                        list1.add(m[curRow1][curCol1]);
                        visited[length - 1 - curRow1][length - 1 - curCol1] = 1;
                        list2.add(0, m[length - 1 - curRow1][length - 1 - curCol1]);
                        curRow1--;
                    }else{
                        break;
                    }
                }
            }
            
            count++;
            tr++;
            tc++;
            dr--;
            dc--;
        }
        
        /**
         * 打印
         */
        System.out.println(list1);
        System.out.println(list2);
        if(length % 2 == 0){
            for (int i = 0; i < list1.size(); i++) {
                System.out.print(list1.get(i) + " ");
            }
            for (int i = 0; i < list2.size(); i++) {
                System.out.print(list2.get(i) + " ");
            }
        }else{
            for (int i = 0; i < list1.size(); i++) {
                System.out.print(list1.get(i) + " ");
            }
            System.out.print(m[(length - 1) / 2][(length - 1) / 2] + " ");
            for (int i = 0; i < list2.size(); i++) {
                System.out.print(list2.get(i) + " ");
            }
        }
    }

}