Leetcode题解剑指 Offer 10- II. 青蛙跳台阶问题

PROBLEM

难度简单

MY ANSWER

与求斐波那契数相同，f(n) = f(n - 1) + f(n - 2)，只不过f(0) = 1, f(1) = 1。时间复杂度O(n)，空间复杂度O(1)。

class Solution {
public:
    int numWays(int n) {
        int a = 1, b = 1, c = 0;
        for(int i = 0; i < n; i++) {
            c = (a + b) % 1000000007;
            a = b;
            b = c;
        }
        return a;
    }
};

BETTER SOLUTION

矩阵快速幂

同样适用于求斐波那契数。时间复杂度O(logn)，O(1)。

使用快速幂计算M的n次幂就可以得到结果。

class Solution {
public:
    vector<vector<long long>> multiply(vector<vector<long long>> &a, vector<vector<long long>> &b) {
        vector<vector<long long>> c(2, vector<long long>(2));
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                c[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j];
            }
        }
        return c;
    }

    vector<vector<long long>> matrixPow(vector<vector<long long>> a, int n) {
        vector<vector<long long>> ret = {{1, 0}, {0, 1}};
        while (n > 0) {
            if ((n & 1) == 1) {
                ret = multiply(ret, a);
            }
            n >>= 1;
            a = multiply(a, a);
        }
        return ret;
    }

    int climbStairs(int n) {
        vector<vector<long long>> ret = {{1, 1}, {1, 0}};
        vector<vector<long long>> res = matrixPow(ret, n);
        return res[0][0];
    }
};

SUMMARY

掌握矩阵快速幂法。