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?
瘋狂收割easy題。。。這題是DP，遞推式是dp[i] = dp[i-1] + dp[i-2]，也就是斐波拉切數列。
為什么？可以這樣想：爬到i-1格再跨1步就能到i，爬到i-2格再跨2格就能到i。
覃超還提出了如果能走1/2/3步，或者走了一步之后下一步距離不能跟前一步相同的情況。
public int climbStairs(int n) {
if (n == 1) return 1;
if (n == 2) return 2;
int[] dp = new int[n+1];
dp[1] = 1;
dp[2] = 2;
for (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
