leecode 解题总结：70. Climbing Stairs

#include <iostream>#include <stdio.h>#include <vector>using namespace std;/*问题: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.分析:n个台阶的楼梯，每次只允许爬一个或者两个，问总共有多少种方式。这个是动态规划设dp[i]表示到达第i层台阶的方式，则dp[i] = dp[i-1] + dp[i-2]边界:dp[1] = 1,dp[2]=2目标：dp[n]dp[3]1,1,12,11,2所以dp[3]共有3种输入:123输出:123*/class Solution {public:    int climbStairs(int n) {if(n <= 0){return 0;}if(n <= 2){return n;}        vector<int> steps(n + 1 , 0);//初始化边界值steps.at(1) = 1;steps.at(2) = 2;for(int i = 3 ; i <= n ; i++){steps.at(i) = steps.at(i-1) + steps.at(i-2);}return steps.at(n);    }};void process(){ int num; Solution solution; while(cin >> num ) { int result = solution.climbStairs(num); cout << result << endl; }}int main(int argc , char* argv[]){process();getchar();return 0;}