Prove safety offer66: range of motion of the robot

1 Title Description

　　A ground grid and m rows n columns. A robot moves from the grid coordinates 0,0, every time only left, right, upper, lower four directions a cell, but can not enter the row and column coordinates of the grid is greater than the sum of the number of bits k. For example, when k is 18, the robot can enter the box (35, 37), because 3 + 3 + 5 + 7 = 18. However, it can not enter the box (35, 38), because 3 + 3 + 5 + 8 = 19. Will the robot be able to reach the number of lattice?

2 ideas and methods

　　Step 1: Create a behavior rows, cols as an array of bool type, used to mark the location walked, initialized to false, true representation through .

　　Step two: function sum (int row, int col) for calculating bit i, j-th bit of the digital sum.

　　The third step: inspection查能否进入坐标为(row, col)的方格

　　Step Four: getSum(row,col) <= threshold && !visited[row*cols+col]表示可以走。

　　Fifth step: a fourth judgment step satisfy, this flag is true grid in the flag, flag gone through.

　　Step Six: recursion on this position, down, left, right, return recursively, down, left, right, plus 1 (add your own) and.

　　Step Seven: movingCount function call movingCountCorefunction, i.e., the initial position of i = 0, j = 0, let the result of the recursive returned directly.

3 C ++ core code　　　　

 1 class Solution {
 2 public:
 3     int movingCount(int threshold, int rows, int cols)
 4     {
 5         if(threshold <= 0|| rows <1 || cols<1)
 6             return 0;
 7 
 8         bool *visited = new bool[rows * cols];
 9         memset(visited,0,rows * cols);
10 
11         intCOUNT = movingCountCore (threshold, rows, cols, 0 , 0 , visited);
 12 is          
13 is          Delete [] visited;
 14          
15          return COUNT;
 16      }
 . 17      
18 is      int movingCountCore ( int threshold, int rows, int cols, int Row, int COL , BOOL * visited)
 . 19      {
 20 is          int COUNT = 0 ;
 21 is          // check can enter the coordinates (row, col) of the box 
22 is          IF (Row> = 0 && row < rows && col>=0 && col<cols && getSum(row,col) <= threshold && !visited[row*cols+col])
23         {
24             visited[row*cols + col] = true;
25             count = 1 + movingCountCore(threshold,rows,cols,row-1,col,visited)
26                     + movingCountCore(threshold,rows,cols,row+1,col,visited)
27                     + movingCountCore(threshold,rows,cols,row,col-1,visited)
28                     + movingCountCore(threshold,rows,cols,row,col+1,visited);
29         }
30         return count;
31     }
32 
33     // 得到数位和
34     int getSum(int row, int col)
35     {
36         int sum = 0;
37         while(row>0){
38             sum += row%10;
39             row /= 10;
40         }
41         while(col>0){
42             sum += col%10;
43             col /= 10;
44         }
45         return sum;
46     }
47 };

View Code

Reference material

https://blog.csdn.net/zjwreal/article/details/89296096，https://blog.csdn.net/u012477435/article/details/83351659#_1782

https://blog.csdn.net/qq_43109561/article/details/89670163

https://blog.csdn.net/Mr_XiaoZ/article/details/81174055?utm_source=blogxgwz1