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935. Knight Dialer

Posted on Edited on In Disqus:
Symbols count in article: 2.8k Reading time ≈ 3 mins.

dp优化矩阵乘法 O(logn) time O(1) space
状态转移矩阵用column-based竖着看才行

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class Solution {
public:
    int knightDialer(int N) {
        vector<vector<long>> mtx = {
            {0, 0, 0, 0, 1, 0, 1, 0, 0, 0},
            {0, 0, 0, 0, 0, 0, 1, 0, 1, 0},
            {0, 0, 0, 0, 0, 0, 0, 1, 0, 1},
            {0, 0, 0, 0, 1, 0, 0, 0, 1, 0},
            {1, 0, 0, 1, 0, 0, 0, 0, 0, 1},
            {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
            {1, 1, 0, 0, 0, 0, 0, 1, 0, 0},
            {0, 0, 1, 0, 0, 0, 1, 0, 0, 0},
            {0, 1, 0, 1, 0, 0, 0, 0, 0, 0},
            {0, 0, 1, 0, 1, 0, 0, 0, 0, 0}
        };
        mtx = pow(mtx, N - 1);
        vector<vector<long>> f{vector<long>(10, 1)};
        f = mul(f, mtx);
        int res = 0;
        for (int i = 0; i < 10; ++i) {
            res = (res + f[0][i]) % M;
        }
        return res;
    }

    vector<vector<long>> pow(vector<vector<long>> &mtx, int N) {
        int n = mtx.size();
        vector<vector<long>> res(n, vector<long>(n));
        for (int i = 0; i < n; ++i) {
            res[i][i] = 1;
        }
        while (N > 0) {
            if (N & 1) {
                res = mul(res, mtx);
            }
            mtx = mul(mtx, mtx);
            N >>= 1;
        }
        return res;
    }

    vector<vector<long>> mul(const vector<vector<long>> &a, const vector<vector<long>> &b) {
        int m = a.size(), n = b.size(), p = b[0].size();
        vector<vector<long>> res(m, vector<long>(p));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < p; ++j) {
                for (int k = 0; k < n; ++k) {
                    res[i][j] = (res[i][j] + a[i][k] * b[k][j]) % M;
                }
            }
        }
        return res;
    }

    int M = 1e9 + 7;
};

dp O(N) time O(1) space
bottom up
最底层是N=1每一层往上迭代累加即可
f[i]表示最后一个数为i的可能的拨号方式有多少种
f[i]表示当前层到最底层从i开始的不同number的个数

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class Solution {
public:
    int knightDialer(int N) {
        vector<vector<int>> next = {{4, 6}, {6, 8}, {7, 9}, {4, 8}, {0, 3, 9}, {}, {0, 1, 7}, {2, 6}, {1, 3}, {2, 4}};
        int M = 1e9 + 7;
        vector<int> f(10, 1);
        for (int i = 1; i < N; ++i) {
            vector<int> t(10);
            t.swap(f);
            for (int u = 0; u < 10; ++u) {
                for (int v : next[u]) {
                    f[v] = (f[v] + t[u]) % M;
                }
            }
        }
        int res = 0;
        for (int i = 0; i < 10; ++i) {
            res = (res + f[i]) % M;
        }
        return res;
    }
};

dfs+memo O(N) time O(N) space

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class Solution {
public:
    int knightDialer(int N) {
        next = {{4, 6}, {6, 8}, {7, 9}, {4, 8}, {0, 3, 9}, {}, {0, 1, 7}, {2, 6}, {1, 3}, {2, 4}};
        m.resize(10, vector<int>(N + 1, -1));
        int res = 0;
        for (int i = 0; i < 10; ++i) {
            res = (res + count(i, N)) % M;
        }
        return res;
    }

    int count(int u, int N) {
        if (N == 1) return 1;
        if (m[u][N] != -1) return m[u][N];
        int res = 0;
        for (int v : next[u]) {
            res = (res + count(v, N - 1)) % M;
        }
        return m[u][N] = res;
    }

    int M = 1e9 + 7;
    vector<vector<int>> m, next;
};