[LeetCode16]Climbing Stairs

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?

DP. The transition function should be:
 f(n) = f(n-1) + f(n-2)  n>2;
     or = 1   n=1
    or  = 2   n=2

Java

public int climbStairs(int n) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        int f1 = 1;int f2 = 2;if(n == 1)return f1;if(n == 2)return f2;int fn = 0;for(int i = 3;i<=n;i++){fn = f1 + f2;f1 = f2;f2 = fn;}return fn;    }

只需要结果，三个变量就够了,可以不用数组

c++

int climbStairs(int n) {        if(n==0) return 0;    if(n==1) return 1;    if(n==2) return 2;    int fn=0, fn_1 = 1,fn_2=2;    for(int i=3;i<=n;i++){        fn = fn_2+fn_1;        fn_1 = fn_2;        fn_2 = fn;    }    return fn;    }