0

我正在尝试用递归解决这个问题。
问题是：对于正整数的二维数组，我如何返回最长路径（步数），以便最长路径的每个单元格中的值来自整数的降序序列，并且每个单元格和单元格之间的差异是给定数字（num）。
假设n是单元格的值，因此 ( n - num) 是正数（非零）。
我不能使用任何循环（for、while、...等）

方法：

public static int longestPath(int matrix[][],int number){...
return(__overloading__);
}

例如：

int number=1;
int [][]matrix ;
       matrix = new int[][]{{8, 15, 20, 33, 35},
                           {60, 59, 58, 32, 31},
                           {59, 17, 57, 56, 55},
                           {55, 15, 13, 58, 16}};

     System.out.print(" longestPath= "+ longestPath(matrix, num));
   }

如果我们搜索差数= 1的最长路径

1-在单元格矩阵[0][3]中，路径长为3，该路径中的值是33 -> 32 -> 31以矩阵[1][4]结尾

2-在单元矩阵 [1][0 ] 中，路径 long 为 6，路径 60 -> 59 -> 58 -> 57 -> 56 -> 55 中的值以矩阵 [2][4]结尾

3-在单元格矩阵[1][0]中，路径长为2，此路径中的值为60 -> 59以矩阵[2][0]结尾

所以该方法必须返回最长的路径，它的 6

如果我们搜索差数= 2的最长路径

1-在单元格矩阵[2][1]中，路径长为3，此路径中的值是17 -> 15 -> 13 以矩阵[3][2]结尾

该方法必须返回其 3 的最长路径。

我的非工作代码：

public class CC {

    public static int longestPath (int arr[][] , int  num){
        return longestPath(arr,arr.length-1,arr[0].length-1,num,0);
    }

    public static int longestPath (int arr[][],int rows,int cols,int num,int max){
        System.out.println("==> longestPath() arr value=" + arr[rows][cols] + " rows:"+rows + " cols:"+cols + " max:"+max);
        if (cols ==0 && rows != 0  ){  
            cols = arr[0].length-1;
            rows--;
        }
        if (rows ==0  && cols==0 ){
            System.out.println("finish");
            return 0;
        }

        int steps = searchPath(arr,rows,cols,num,max);
        if (steps > max) max=steps;
        longestPath(arr,rows,cols-1,num,max);
        return max ;
    }

    public static int searchPath(int arr[][],int rows,int cols,int num ,int counter){ 
        System.out.println("searchPath() arr value=" + arr[rows][cols] + " rows:"+rows + " cols:"+cols);
        int left=1,right=1,up=1,down=1;

        if ((cols != 0) && arr[rows][cols] - num == arr[rows-1][cols]  ){ // checking up cell
            counter++;
            up = searchPath(arr,rows-1,cols,num,counter);

        }
        if ((rows != arr.length-1)   &&   arr[rows][cols] - num == arr[rows+1][cols] ){ // checking down cell
            counter++;
            down = searchPath(arr,rows+1,cols,num,counter);
            // return counter;
        }
        if ((cols != 0)  &&  arr[rows][cols] - num == arr[rows][cols-1]){ // checking left cell
            counter++;
            left = searchPath(arr,rows,cols-1,num,counter);
            //return counter;
        } 
        if ((cols != arr[0].length-1)   &&   arr[rows][cols] - num == arr[rows][cols+1]  ){ //checking right cell
            counter++;
            right = searchPath(arr,rows,cols+1,num ,counter);
            //return counter;
        } 

        if ((left > right) && (left > up)    && (left > down)) // if left cell is bigger than all other direction return left
            return left;
        if ((right > left) && (right > up)   && (right > down))
            return right;
        if ((down > up)    && (down > right) &&( down > left))
            return down;
        if ((up> down) && (up > right) && (up>left))
            return up;

        return 0;
    }

}

在编写代码时，我遇到了很多运行问题

我做错了什么？提前致谢

4

1 回答 1

0

在发布的解决方案（以及问题中）中，您从左上角（0,0）开始迭代整个二维数组，并希望检查每个元素的邻居。
要检查所有邻居，检查右下邻居以覆盖所有1就足够了：

在此处输入图像描述

1：如果您对向上或向左的下行路径感兴趣，则需要检查4 个方向。这是因为您的图表是有向的。例如，一条从左到右可能无效的路径，从右到左可能是有效的。

通过这样做，您可以简化代码，并减少对邻居的重复测试。
另一种使代码更易于阅读、调试和维护的技术是将其分解为定义明确的简单方法，例如：

private static boolean isValidAddress(int row, int col, int maxRow, int maxCol) {
    if(row < 0 || col < 0) return false;
    if(row >= maxRow || col >= maxCol) return false;
    return true;
}

尝试以下解决方案：

public class Main {

    //moving in two directions, right and down, is sufficient
    //to cover a whole matrix without visiting the same address twice

    public static void main(String[] args) {
        int delta= 1;
        int [][]matrix =  new int[][]{
                                     {8, 15, 20, 33, 35},
                                     {60, 59, 58, 32, 31},
                                     {59, 17, 57, 56, 55},
                                     {55, 15, 13, 58, 16}};
        System.out.print(" longest Path= "+ longestPath(matrix, delta));
    }

     public static int longestPath (int arr[][] , int delta){
            return longestPath(arr, 0, 0, delta , 0);
     }

     //check all matrix elements, keep longest path found
     public static int longestPath (int arr[][],int row,int col, int num, int max){

        int steps = searchPath(arr,row,col,num, 1); //Initial path length is always 1
        if (steps > max) {  max=steps;  }

        if (row == arr.length-1 && col == arr[row].length -1 )  return max;
        col ++;
        if(col == arr[row].length){//end of row exceeded
            row++;    //new row
            col = 0;  //first column
        }

        return longestPath(arr,row,col,num,max);
    }

    public static int searchPath(int arr[][],int row,int col,int num ,int pathLength){

        int[][] neighbors = getNeighbors(arr, row, col, num);
        int rightPath = 0 , downPath = 0;
        //right neighbor
        if(neighbors[0] != null){
            rightPath = searchPath(arr, neighbors[0][0], neighbors[0][1], num, pathLength+1);
        }

        //down neighbor
        if(neighbors[1] != null){
            downPath = searchPath(arr, neighbors[1][0], neighbors[1][1], num, pathLength+1);
        }

        int returnPath = Math.max(rightPath, downPath); //max return value 
        return Math.max(pathLength, returnPath) ; //max of path length and returned value 
    }

    //return neighbors with value smaller by delta
    //returned array[0] represents right neighbor row, col  or null
    //returned array[1] represents down  neighbor row, col  or null
    private static int[][] getNeighbors(int[][] arr, int row, int col, int delta) {

        //moving in two directions, right and down, is sufficient
        //to cover a whole matrix without visiting the same address twice

        int[][] neighbors = {null, null};
        //right neighbor
        int newCol = col +1;
        if(isValidAddress(row, newCol, arr.length, arr[0].length)){
            if(arr[row][col] - arr[row][newCol] == delta){
                neighbors[0] = new int[]{row, newCol};
            }
        }

        //down neighbor
        int newRow = row + 1 ;
        if(isValidAddress(newRow, col, arr.length, arr[0].length)){
            if(arr[row][col] - arr[newRow][col] == delta){
                neighbors[1] = new int[]{newRow, col};
            }
        }

        return neighbors;
    }

    private static boolean isValidAddress(int row, int col, int maxRow, int maxCol) {
        if(row < 0 || col < 0) return false;
        if(row >= maxRow || col >= maxCol) return false;
        return true;
    }
}
于 2019-01-14T18:17:40.337 回答