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?
解析:本题目主要考虑递归和DP两种思想;但是递归的时间太长,大数时太耗时间,故采用DP的思想;利用一个含有3个整形变量的数组进行不断的操作,节省空间和时间的消耗;
代码如下:
int climbStairs(int n) {
int s[]={0,1,2};
if(n<=2) return s[n];
int num=2;
while(num++<n){
s[num%3]=s[(num+1)%3]+s[(num+2)%3];
}
return s[n%3];
}
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