70. Climbing Stairs

题目：

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?

链接： http://leetcode.com/problems/climbing-stairs/

斐波那契数列的应用

一刷，用dp做法，注意在建数组时需要明确长度，行8， 否则行12会报错。

class Solution(object):
    def climbStairs(self, n):

        if n == 1:
            return 1
        elif n == 2:
            return 2
        methods = [0 for i in range(n)]
        methods[:1] = [1, 2]
        
        for index in range(2, n):
            methods[index] = methods[index - 1] + methods[index - 2]
        return methods[n - 1]

用2个变量表示，节省空间复杂度

class Solution(object):
    def climbStairs(self, n):

        if n == 1:
            return 1
        elif n == 2:
            return 2
        first = 1
        second = 2
        
        for i in range(2, n):
            current = first + second
            first, second = second, current
        
        return current

简化初始判断

class Solution(object):
    def climbStairs(self, n):
        first, second = 1, 2
        for i in range(2, n + 1):
            first, second = second, first + second
        return first

矩阵解法，未懂：

class Solution(object):
    def matrix_multiply(self, M, N):
        a = M[0][0] * N[0][0] + M[0][1] * N[1][0]
        b = M[0][0] * N[0][1] + M[0][1] * N[1][1]
        c = M[1][0] * N[0][0] + M[1][1] * N[1][0]
        d = M[1][0] * N[0][1] + M[1][1] * N[1][1]
        return [[a, b], [c, d]]
    
    def climbStairs(self, n):
        if n == 1:
            return 1
        
        ret = [[1, 0], [0, 1]]
        test_matrix = [[1, 1], [1, 0]]
        while n:
            if n % 2:
                ret = self.matrix_multiply(ret, test_matrix)
            n /= 2
            test_matrix = self.matrix_multiply(test_matrix, test_matrix)

        return ret[0][0]

2/14/2017, Java, 斐波那契数列解法

public class Solution {
    public int climbStairs(int n) {
        if (n <= 2) return n;
        int[] ret = new int[n];
        ret[0] = 1;
        ret[1] = 2;
        
        for(int i = 2; i < n; i++) {
            ret[i] = ret[i-1] + ret[i-2];
        }
        return ret[n-1];
    }
}