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[LintCode] Climbing Stairs

jemygraw / 3053人閱讀

摘要：無需動(dòng)規(guī)，無需額外空間，等同于菲波那切數(shù)列。當(dāng)然嚕，也可以動(dòng)規(guī)，記住就好。

Problem

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?

Example

Given an example n=3 , 1+1+1=2+1=1+2=3

return 3

Note

無需動(dòng)規(guī)，無需額外空間，等同于菲波那切數(shù)列。
當(dāng)然嚕，也可以動(dòng)規(guī)，記住dp[0] = 1就好。

Solution Fibonacci 
public class Solution {
    public int climbStairs(int n) {
        if (n < 2) return 1;
        int one = 1, two = 1, total = 0;
        for (int i = 2; i <= n; i++) {
            res = one + two;
            two = one;
            one = res;
        }
        return res;
    }
}
DP 
public class Solution {
    public int climbStairs(int n) {
        int[] dp = new int[n+1];
        dp[0] = 1;
        dp[1] = 1;
        for (int i = 2; i <= n; i++) dp[i] = dp[i-1] + dp[i-2];
        return dp[n];
    }
}

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