题目链接: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?
这道题的要求是爬n阶楼梯,每次只可以爬1步或2步,总共有多少种不同方法能爬到顶?
动态规划,假设要爬到第i阶,可以由i-1和i-2阶1次过去,因此dp[i] = dp[i-1] + dp[i-2]。
可以注意到,爬到n阶台阶的方案数目其实就是求斐波那契数列的第n+1位。
时间复杂度:O(n)
空间复杂度:O(n)
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class Solution
{
public:
int climbStairs(int n)
{
vector<int> v(n + 1);
v[0] = v[1] = 1;
for(int i = 2; i <= n; ++ i)
v[i] = v[i - 1] + v[i - 2];
return v[n];
}
};
由于动态规划的时候只用到了前面2步,因此可以用两个变量记录一下,这样就将空间复杂度降到O(1)了。
时间复杂度:O(n)
空间复杂度:O(1)
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class Solution
{
public:
int climbStairs(int n)
{
int n2 = 0, n1 = 1, res = 0;
for(int i = 0; i < n; ++ i)
{
res = n2 + n1;
n2 = n1;
n1 = res;
}
return res;
}
};