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  • LeetCode_70. Climbing Stairs

    70. Climbing Stairs

    Easy

    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?

    Note: Given n will be a positive integer.

    Example 1:

    Input: 2
    Output: 2
    Explanation: There are two ways to climb to the top.
    1. 1 step + 1 step
    2. 2 steps
    

    Example 2:

    Input: 3
    Output: 3
    Explanation: There are three ways to climb to the top.
    1. 1 step + 1 step + 1 step
    2. 1 step + 2 steps
    3. 2 steps + 1 step
    package leetcode.easy;
    
    public class ClimbingStairs {
    	@org.junit.Test
    	public void test() {
    		int n1 = 2;
    		int n2 = 3;
    		System.out.println(climbStairs1(n1));
    		System.out.println(climbStairs1(n2));
    		System.out.println(climbStairs2(n1));
    		System.out.println(climbStairs2(n2));
    		System.out.println(climbStairs3(n1));
    		System.out.println(climbStairs3(n2));
    		System.out.println(climbStairs4(n1));
    		System.out.println(climbStairs4(n2));
    		System.out.println(climbStairs5(n1));
    		System.out.println(climbStairs5(n2));
    		System.out.println(climbStairs6(n1));
    		System.out.println(climbStairs6(n2));
    	}
    
    	public int climbStairs1(int n) {
    		return climb_Stairs(0, n);
    	}
    
    	public int climb_Stairs(int i, int n) {
    		if (i > n) {
    			return 0;
    		}
    		if (i == n) {
    			return 1;
    		}
    		return climb_Stairs(i + 1, n) + climb_Stairs(i + 2, n);
    	}
    
    	public int climbStairs2(int n) {
    		int[] memo = new int[n + 1];
    		return climb_Stairs(0, n, memo);
    	}
    
    	public int climb_Stairs(int i, int n, int memo[]) {
    		if (i > n) {
    			return 0;
    		}
    		if (i == n) {
    			return 1;
    		}
    		if (memo[i] > 0) {
    			return memo[i];
    		}
    		memo[i] = climb_Stairs(i + 1, n, memo) + climb_Stairs(i + 2, n, memo);
    		return memo[i];
    	}
    
    	public int climbStairs3(int n) {
    		if (n == 1) {
    			return 1;
    		}
    		int[] dp = new int[n + 1];
    		dp[1] = 1;
    		dp[2] = 2;
    		for (int i = 3; i <= n; i++) {
    			dp[i] = dp[i - 1] + dp[i - 2];
    		}
    		return dp[n];
    	}
    
    	public int climbStairs4(int n) {
    		if (n == 1) {
    			return 1;
    		}
    		int first = 1;
    		int second = 2;
    		for (int i = 3; i <= n; i++) {
    			int third = first + second;
    			first = second;
    			second = third;
    		}
    		return second;
    	}
    
    	public int climbStairs5(int n) {
    		int[][] q = { { 1, 1 }, { 1, 0 } };
    		int[][] res = pow(q, n);
    		return res[0][0];
    	}
    
    	public int[][] pow(int[][] a, int n) {
    		int[][] ret = { { 1, 0 }, { 0, 1 } };
    		while (n > 0) {
    			if ((n & 1) == 1) {
    				ret = multiply(ret, a);
    			}
    			n >>= 1;
    			a = multiply(a, a);
    		}
    		return ret;
    	}
    
    	public int[][] multiply(int[][] a, int[][] b) {
    		int[][] c = new int[2][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;
    	}
    
    	public int climbStairs6(int n) {
    		double sqrt5 = Math.sqrt(5);
    		double fibn = Math.pow((1 + sqrt5) / 2, n + 1) - Math.pow((1 - sqrt5) / 2, n + 1);
    		return (int) (fibn / sqrt5);
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/denggelin/p/11589805.html
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