Climbing Stairs

最新推荐文章于 2024-01-18 13:59:59 发布

最新推荐文章于 2024-01-18 13:59:59 发布 · 60 阅读

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本文探讨了经典的爬楼梯问题，提供了两种不同的算法解决方案。一种是通过迭代的方法逐步计算到达楼梯顶部的不同方式的数量；另一种则利用矩阵快速幂的方式进行高效求解。

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

public class Solution {
    public int climbStairs(int n) {
        if (n < 1) {
        	return 0;
        }
        if (n == 1 || n == 2) {
        	return n;
        }
        int res = 2;
        int pre = 1;
        int tmp = 0;
        for (int i = 3; i <= n; i++) {
			tmp = res;
			res = res + pre;
			pre = tmp;
		}
        return res;
    }
}

public class Solution {
    public int climbStairs(int n) {
        if (n < 1) {
        	return 0;
        }
        if (n == 1 || n == 2) {
        	return n;
        }
        int[][] base = {{1, 1}, {1, 0}};
        int[][] res = help(base, n - 2);
        return 2 * res[0][0] + res[1][0];
    }

	private int[][] help(int[][] m, int p) {
		// TODO Auto-generated method stub
		int[][] res = new int[m.length][m[0].length];
		for (int i = 0; i < res.length; i++) {
			res[i][i] = 1;
		}
		int[][] tmp = m;
		for (; p != 0; p >>= 1) {
			if ((p & 1) != 0) {
				res = help2(res, tmp);
			}
			tmp = help2(tmp, tmp);
		}
		return res;
	}

	private int[][] help2(int[][] m1, int[][] m2) {
		// TODO Auto-generated method stub
		int[][] res = new int[m1.length][m2[0].length];
		for (int i = 0; i < m2[0].length; i++) {
			for (int j = 0; j < m1.length; j++) {
				for (int k = 0; k < m2.length; k++) {
					res[i][j] += m1[i][k] * m2[k][j];
				}
			}
		}
		return res;
	}
}