机器人的运动范围

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?

和之前矩阵中的路径解题思路差不多, 区别就是使用一个出参

class Solution {
public:
    int numSum(int num1, int num2) {    // 计算位数之和
        int res = 0;
        while (num1 % 10) {
            res += num1 % 10;
            num1 /= 10;
        }
 
        while (num2 % 10) {
            res += num2 % 10;
            num2 /= 10;
        }
        return res;
    }
     
    void movingCount(int threshold, int rows, int cols, int row, int col, bool *visited, int &len) {
        int sum = numSum(col, row);
        if (sum > threshold)
            return;
 
        if ((row >= 0) && (row < rows) && (col >=0) && (col < cols) // 定边界
            && (visited[row * cols + col] == false)) {          // 没有访问过
             
            len++;      // 长度+1
            visited[row * cols + col] = true;
            movingCount(threshold, rows, cols, row + 1, col, visited, len);
            movingCount(threshold, rows, cols, row, col + 1, visited, len);
            movingCount(threshold, rows, cols, row - 1, col, visited, len);
            movingCount(threshold, rows, cols, row, col - 1, visited, len);
        }
 
    }
 
    int movingCount(int threshold, int rows, int cols)
    {
        if ((threshold < 0) || (rows < 1) || (cols < 1))
            return 0;
 
        bool *visited = new bool[rows * cols];      // 判断单元格是否被访问过
        memset(visited, 0, rows * cols);
 
        int len = 0;
        movingCount(threshold, rows, cols, 0, 0, visited, len);
        delete[] visited;
        
        return len;
    }
};

书上版本, 感觉差不多, 只是分了更多的模块

class Solution {
public:
    int getDigitSum(int number) {
        int sum = 0;
        while (number > 0) {
            sum += number % 10;
            number /= 10;
        }
        return sum;
    }
    
    bool check(int threshold, int rows, int cols, int row, int col, bool *visited) {
        if ((row >= 0) && (row < rows) && (col >= 0) && (col < cols)
            && (getDigitSum(row) + getDigitSum(col) <= threshold)
            && (!visited[row * cols + col]))
            return true;
        else 
            return false;
    }

    int movingCountCore(int threshold, int rows, int cols, int row, int col, bool *visited) {
        int count = 0;
        if (check(threshold, rows, cols, row, col, visited)) {
            visited[row * cols + col] = true;

            count = 1 + movingCountCore(threshold, rows, cols, row + 1, col, visited)
                      + movingCountCore(threshold, rows, cols, row, col + 1, visited)
                      + movingCountCore(threshold, rows, cols, row - 1, col, visited)
                      + movingCountCore(threshold, rows, cols, row, col - 1, visited);
                    //(int threshold, int rows, int cols, int row, int col, bool *visited)        
        }
        return count;
    }

    int movingCount(int threshold, int rows, int cols)
    {
        if ((threshold < 0) || (rows <= 0) || (cols <= 0))
            return 0;

        bool *visited = new bool[rows * cols];
        memset(visited, 0, rows * cols);

        int count = movingCountCore(threshold, rows, cols, 0, 0, visited);
        delete[] visited;

        return count;
    }
};
posted @ 2019-03-22 13:49  张飘扬  阅读(91)  评论(0编辑  收藏  举报