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  • 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;
        }
    };
    
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  • 原文地址:https://www.cnblogs.com/hesper/p/10577769.html
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