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576. 出界的路径数

给你一个大小为 m x n 的网格和一个球。球的起始坐标为 [startRow, startColumn] 。你可以将球移到在四个方向上相邻的单元格内(可以穿过网格边界到达网格之外)。你 最多 可以移动 maxMove 次球。

给你五个整数 mnmaxMovestartRow 以及 startColumn ,找出并返回可以将球移出边界的路径数量。因为答案可能非常大,返回对 109 + 7 取余 后的结果。

 

示例 1:

输入:m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0
输出:6

示例 2:

输入:m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1
输出:12

 

提示:

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class Solution { public: int findPaths(int m, int n, int maxMove, int startRow, int startColumn) { } };

python3 解法, 执行用时: 100 ms, 内存消耗: 16.3 MB, 提交时间: 2023-10-24 07:52:15

class Solution:
    def findPaths1(self, m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int:
        MOD = 10**9 + 7

        outCounts = 0
        dp = [[[0] * n for _ in range(m)] for _ in range(maxMove + 1)]
        dp[0][startRow][startColumn] = 1
        for i in range(maxMove):
            for j in range(m):
                for k in range(n):
                    if dp[i][j][k] > 0:
                        for j1, k1 in [(j - 1, k), (j + 1, k), (j, k - 1), (j, k + 1)]:
                            if 0 <= j1 < m and 0 <= k1 < n:
                                dp[i + 1][j1][k1] = (dp[i + 1][j1][k1] + dp[i][j][k]) % MOD
                            else:
                                outCounts = (outCounts + dp[i][j][k]) % MOD
        
        return outCounts
        

    def findPaths(self, m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int:
        MOD = 10**9 + 7

        outCounts = 0
        dp = [[0] * n for _ in range(m)]
        dp[startRow][startColumn] = 1
        for i in range(maxMove):
            dpNew = [[0] * n for _ in range(m)]
            for j in range(m):
                for k in range(n):
                    if dp[j][k] > 0:
                        for j1, k1 in [(j - 1, k), (j + 1, k), (j, k - 1), (j, k + 1)]:
                            if 0 <= j1 < m and 0 <= k1 < n:
                                dpNew[j1][k1] = (dpNew[j1][k1] + dp[j][k]) % MOD
                            else:
                                outCounts = (outCounts + dp[j][k]) % MOD
            dp = dpNew
        
        return outCounts

java 解法, 执行用时: 4 ms, 内存消耗: 39.8 MB, 提交时间: 2023-10-24 07:51:45

class Solution {
    public int findPaths1(int m, int n, int maxMove, int startRow, int startColumn) {
        final int MOD = 1000000007;
        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        int outCounts = 0;
        int[][][] dp = new int[maxMove + 1][m][n];
        dp[0][startRow][startColumn] = 1;
        for (int i = 0; i < maxMove; i++) {
            for (int j = 0; j < m; j++) {
                for (int k = 0; k < n; k++) {
                    int count = dp[i][j][k];
                    if (count > 0) {
                        for (int[] direction : directions) {
                            int j1 = j + direction[0], k1 = k + direction[1];
                            if (j1 >= 0 && j1 < m && k1 >= 0 && k1 < n) {
                                dp[i + 1][j1][k1] = (dp[i + 1][j1][k1] + count) % MOD;
                            } else {
                                outCounts = (outCounts + count) % MOD;
                            }
                        }
                    }
                }
            }
        }
        return outCounts;
    }


    public int findPaths(int m, int n, int maxMove, int startRow, int startColumn) {
        final int MOD = 1000000007;
        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        int outCounts = 0;
        int[][] dp = new int[m][n];
        dp[startRow][startColumn] = 1;
        for (int i = 0; i < maxMove; i++) {
            int[][] dpNew = new int[m][n];
            for (int j = 0; j < m; j++) {
                for (int k = 0; k < n; k++) {
                    int count = dp[j][k];
                    if (count > 0) {
                        for (int[] direction : directions) {
                            int j1 = j + direction[0], k1 = k + direction[1];
                            if (j1 >= 0 && j1 < m && k1 >= 0 && k1 < n) {
                                dpNew[j1][k1] = (dpNew[j1][k1] + count) % MOD;
                            } else {
                                outCounts = (outCounts + count) % MOD;
                            }
                        }
                    }
                }
            }
            dp = dpNew;
        }
        return outCounts;
    }
}

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