LeetCode-Climbing Stairs（爬楼梯问题）

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?

第一反应，递归求解，貌似很简单。但是不幸，超时

public int climbStairs1(int n) {
    	if (n == 1 || n == 2) {
    		return n;
    	}
    	return climbStairs1(n-1) + climbStairs1(n-2);
    }

如果所示，标颜色的需要重复递归调用计算，所以动态规划的思想就是把重叠子问题存储下来，下次调用直接查表即可。比如在第一次递归调用时，n=3已经将结果计算出来，在n=4的时候就不需要就算了，上代码：

public int climbStairs(int n) {
    	if (n == 0 || n == 1 || n == 2) {
    		return n;
    	}
        int[] r = new int[n+1];
        r[1] = 1;
        r[2] = 2;
        for (int i = 3; i <= n; i++) {
        	r[i] = r[i-1] + r[i-2];
        }
        return r[n];
    }