[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;
    }