[LeetCode] Climbing Stairs
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[Problem]
[Analysis]
Fibnocci Sequency. We can climb n-2 stairs end with 2 stairs orclimb n-1 stairs end with 1 stair to reach n stairs. That meansf(n) = f(n-2) + f(n-1).
In case of TLE for a large n, we should not use a recursive style for thisprogram.
[Solution]
说明:版权所有,转载请注明出处。Coder007的博客
You are climbing a stair case. It takes n steps to reachto the top.
Each time you can either climb 1 or 2 steps. In how manydistinct ways can you climb to the top?
[Analysis]
Fibnocci Sequency. We can climb n-2 stairs end with 2 stairs orclimb n-1 stairs end with 1 stair to reach n stairs. That meansf(n) = f(n-2) + f(n-1).
In case of TLE for a large n, we should not use a recursive style for thisprogram.
[Solution]
class Solution {
public:
int climbStairs(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int fab[n+1];
fab[0] = 1;
fab[1] = 1;
for(int i = 2; i <= n; ++i){
fab[i] = fab[i-1] + fab[i-2];
}
return fab[n];
}
};
说明:版权所有,转载请注明出处。Coder007的博客
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