面试题67 机器人的运动范围

题目描述

地上有一个m行和n列的方格。一个机器人从坐标0,0的格子开始移动,每一次只能向左,右,上,下四个方向移动一格,但是不能进入行坐标和列坐标的数位之和大于k的格子。 例如,当k为18时,机器人能够进入方格(35,37),因为3+5+3+7 = 18。但是,它不能进入方格(35,38),因为3+5+3+8 = 19。请问该机器人能够达到多少个格子?
 1 class Solution {
 2 public:
 3     int sum(int row, int col){
 4         int num = 0;
 5         while (row){
 6             num += row % 10;
 7             row /= 10;
 8         }
 9         while (col){
10             num += col % 10;
11             col /= 10;
12         }
13         return num;
14     }
15     void m(int threshold, int rows, int cols, int i, int j, int &count, int isWalked[]){
16         if (i - 1 >= 0 && isWalked[(i - 1) * cols + j] == 0 && sum(i - 1, j) <= threshold){
17             isWalked[(i - 1) * cols + j] = 1;
18             count++;
19             m(threshold, rows, cols, i - 1, j, count, isWalked);
20         }
21         if (i + 1 < rows && isWalked[(i + 1) * cols + j] == 0 && sum(i + 1, j) <= threshold){
22             isWalked[(i + 1) * cols + j] = 1;
23             count++;
24             m(threshold, rows, cols, i + 1, j, count, isWalked);
25         }
26         if (j - 1 >= 0 && isWalked[i * cols + j - 1] == 0 && sum(i, j - 1) <= threshold){
27             isWalked[i * cols + j - 1] = 1;
28             count++;
29             m(threshold, rows, cols, i, j - 1, count, isWalked);
30         }
31         if (j + 1 < cols && isWalked[i * cols + j + 1] == 0 && sum(i, j + 1) <= threshold){
32             isWalked[i * cols + j + 1] = 1;
33             count++;
34             m(threshold, rows, cols, i, j + 1, count, isWalked);
35         }
36     }
37     int movingCount(int threshold, int rows, int cols)
38     {
39         if (threshold < 0)
40             return 0;
41         int i = 0, j = 0, count = 1;
42         int *isWalked = new int[rows * cols];
43         isWalked[0] = 1;
44         for (int i = 1; i < rows * cols; i++)
45             isWalked[i] = 0;
46         m(threshold, rows, cols, i, j, count, isWalked);
47         return count;
48     }
49 };

 

 

posted @ 2016-04-10 22:11  早杰  阅读(380)  评论(0编辑  收藏  举报