題目:
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?
難度:easy 通過率:38.3%
這道題題意是說爬樓梯有一次爬一格和一次爬兩格兩種方法,問爬n階樓梯有多少方案?一開始直接使用遞歸,每次爬法都等於上一次是爬兩格或者上一次爬一次的爬法運用加法原理加起來,代碼如下:
class Solution {
public:
int climbStairs(int n) {
if(n == 1 ) return 1;
if(n == 2) return 2;
return climbStairs(n-1) +climbStairs(n-2);
}
};
然而超時了~於是運用動態轉移,想法和遞歸差不多,得代碼如下:
時間複雜度:O(n)
class Solution {
public:
int climbStairs(int n) {
if(n == 1 ) return 1;
if(n == 2) return 2;
vector<int> v(n);
v[0] = 1;
v[1] = 2;
for(int i = 2; i < n; i++) {
v[i] = v[i-1] + v[i-2];
}
return v[n-1];
}
};