70. Climbing Stairs
2017-01-15 14:11
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problem:
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?
Note: Given n will be a positive integer.
本问题是一个计算n层阶梯情况下,走到顶端的路径种数(要求每次只能上1层或者2层阶梯)。
这是一个动态规划的题目:
n = 1 时 ways[1] = 1;
n = 2 时 ways[2] = 2;
n = 3 时 ways[3] = 3;
…
n = k 时 ways[K] = ways[k-1] + ways[k-2];
明显的,这是著名的斐波那契数列问题
class Solution {
public:
int climbStairs(int n) {
//如果0层,返回0
if (n <= 0)
return 0;
//如果只有1层阶梯,则只有一种方式
else if (n == 1)
return 1;
//若有两层阶梯,则有两种方式(每次走一层,一次走两层)
else if (n == 2)
return 2;
int *r = new int
;
//其余的情况方式总数 = 最终剩余1层的方式 + 最终剩余两层阶梯的方式
r[0] = 1;
r[1] = 2;
for (int i = 2; i < n; i++)
r[i] = r[i - 1] + r[i - 2];
int ret = r[n - 1];
delete []r;
return ret;
}
};
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?
Note: Given n will be a positive integer.
本问题是一个计算n层阶梯情况下,走到顶端的路径种数(要求每次只能上1层或者2层阶梯)。
这是一个动态规划的题目:
n = 1 时 ways[1] = 1;
n = 2 时 ways[2] = 2;
n = 3 时 ways[3] = 3;
…
n = k 时 ways[K] = ways[k-1] + ways[k-2];
明显的,这是著名的斐波那契数列问题
class Solution {
public:
int climbStairs(int n) {
//如果0层,返回0
if (n <= 0)
return 0;
//如果只有1层阶梯,则只有一种方式
else if (n == 1)
return 1;
//若有两层阶梯,则有两种方式(每次走一层,一次走两层)
else if (n == 2)
return 2;
int *r = new int
;
//其余的情况方式总数 = 最终剩余1层的方式 + 最终剩余两层阶梯的方式
r[0] = 1;
r[1] = 2;
for (int i = 2; i < n; i++)
r[i] = r[i - 1] + r[i - 2];
int ret = r[n - 1];
delete []r;
return ret;
}
};
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