LeetCode Climbing Stairs

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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?

思路分析：考察DP，定义ClimbWays数组，ClimbWays[n]表示n steps不同的走法，如果第n步是一个单步，这种情况的走法有ClimbWays[n-1]种；如果第n步是一个双步，这种情况的走法有ClimbWays[n-2]种；因此可以得到DP方程ClimbWays[n]=ClimbWays[n-1] + ClimbWays[n-2] .

public class Solution {
    public int climbStairs(int n) {
        //04:10
        int [] climbWays = new int[n+1];
        if(n == 0) return 0;
        if(n == 1) return 1;
        climbWays[1] = 1;
        climbWays[2] = 2;
        
        for(int i = 3; i <= n; i++){
            climbWays[i] = climbWays[i-1] + climbWays[i-2];
        }
        return climbWays[n];
    }
    //04:14
}

LeetCode Climbing Stairs

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原文地址：http://blog.csdn.net/yangliuy/article/details/41591527