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?
class Solution {
public:
int climbStairs(int n) {
if (n <= 1) return 1;
vector<int> dp(n);
dp[0] = 1; //0 代表第一层 因为i＝n就退出forloop了
dp[1] = 2; // 上第二层有两种方式 第一种是1次2step，第二种是2次1step
for (int i = 2; i < n; ++i) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp.back();
}
};
int climbStairs(int n) {
int a = 1, b = 1;
while (n--)
a = (b += a) - a;
return a;
}
class Solution {
public:
int climbStairs(int n) {
int a=1;
int b=2;
if(n==1) return a;
if (n==2) return b;
while(--n){
b+=a;
a=b-a;
}
return a;
}
};
Variable a tells you the number of ways to reach the current step, and b tells you the number of ways to reach the next step. So for the situation one step further up, the old b becomes the new a, and the new b is the old a+b, since that new step can be reached by climbing 1 step from what b represented or 2 steps from what a represented.
REF: