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?
DP启蒙啊，做完这题终于对DP有个大概的认识了，自底向上检表，建完也就ok了，复杂度是O(n)
void iter(int acc, int n, vector<int> &table) { if (acc < 0) return; if (acc == n+1) return; int a = acc-1 < 0? 0 : acc-1; int b = acc-2 <0? 0 : acc-2; table.push_back(table[a] + table[b]); iter(acc+1, n, table); } int climbStairs(int n) { vector<int> table; table.push_back(0); table.push_back(1); table.push_back(2); if (n < 3) return table[n]; iter(3, n,table); return table[n]; }
另附纯递归版，当然是TLE的
int climbstairs_TLE(int n) { if (n == 1) return 1; if (n == 2) return 2; return climbstairs_TLE(n-1) + climbstairs_TLE(n-2); }
原文：http://www.cnblogs.com/agentgamer/p/4052988.html
