《剑指offer》第十三题:机器人的运动范围
// 面试题13:机器人的运动范围 // 题目:地上有一个m行n列的方格。一个机器人从坐标(0, 0)的格子开始移动,它 // 每一次可以向左、右、上、下移动一格,但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。 // 但它不能进入方格(35, 38),因为3+5+3+8=19。请问该机器人能够到达多少个格子? #include <cstdio> int movingCountCore(int threshold, int rows, int cols, int row, int col, bool* visited); bool check(int threshold, int rows, int cols, int row, int col, bool* visited); int getDigitSum(int number); int movingCount(int threshold, int rows, int cols) { //鲁棒性测试 if (threshold < 0 || rows <= 0 || cols <= 0) return 0; //访问路径矩阵 bool* visited = new bool[rows * cols]; for (int i = 0; i < rows * cols; ++i) visited[i] = false; int count = movingCountCore(threshold, rows, cols, 0, 0, visited); delete[] visited; return count; } int movingCountCore(int threshold, int rows, int cols, int row, int col, bool* visited) { int count = 0; //检查该位置是否可以进入 if (check(threshold, rows, cols, row, col, visited)) { visited[row * cols + col] = true; //检查上下左右四个位置是否可以进入 count = 1 + movingCountCore(threshold, rows, cols, row, col - 1, visited) + movingCountCore(threshold, rows, cols, row - 1, col, visited) + movingCountCore(threshold, rows, cols, row, col + 1, visited) + movingCountCore(threshold, rows, cols, row + 1, col, visited); } return count; } bool check(int threshold, int rows, int cols, int row, int col, bool* visited) { //检查row和col位之和是否大于threshold if (row >= 0 && row < rows && col >= 0 && col < cols && getDigitSum(row) + getDigitSum(col) <= threshold && !visited[row * cols + col]) return true; return false; } int getDigitSum(int number) { int sum = 0; while (number > 0) { sum += number % 10; number = number / 10; } return sum; }
// ====================测试代码==================== void test(const char* testName, int threshold, int rows, int cols, int expected) { if (testName != nullptr) printf("%s begins: ", testName); if (movingCount(threshold, rows, cols) == expected) printf("Passed.\n"); else printf("FAILED.\n"); } // 方格多行多列 void test1() { test("Test1", 5, 10, 10, 21); } // 方格多行多列 void test2() { test("Test2", 15, 20, 20, 359); } // 方格只有一行,机器人只能到达部分方格 void test3() { test("Test3", 10, 1, 100, 29); } // 方格只有一行,机器人能到达所有方格 void test4() { test("Test4", 10, 1, 10, 10); } // 方格只有一列,机器人只能到达部分方格 void test5() { test("Test5", 15, 100, 1, 79); } // 方格只有一列,机器人能到达所有方格 void test6() { test("Test6", 15, 10, 1, 10); } // 方格只有一行一列 void test7() { test("Test7", 15, 1, 1, 1); } // 方格只有一行一列 void test8() { test("Test8", 0, 1, 1, 1); } // 机器人不能进入任意一个方格 void test9() { test("Test9", -10, 10, 10, 0); } int main(int agrc, char* argv[]) { test1(); test2(); test3(); test4(); test5(); test6(); test7(); test8(); test9(); return 0; }
分析:递归思想。
class Solution { public: int movingCount(int threshold, int rows, int cols) { if (threshold < 0 || rows <= 0 || cols <= 0) return 0; bool* visited = new bool[rows * cols]; for (int i = 0; i < rows * cols; ++i) visited[i] = false; int count = movingCountCore(threshold, rows, cols, 0, 0, visited); delete[] visited; return count; } 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, col - 1, visited) + movingCountCore(threshold, rows, cols, row - 1, col, visited) + movingCountCore(threshold, rows, cols, row, col + 1, visited) + movingCountCore(threshold, rows, cols, row + 1, col, visited); } return count; } int 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; return false; } int getDigitSum(int number) { int sum = 0; while (number > 0) { sum += number % 10; number /= 10; } return sum; } };