【LeetCode】Climbing Stairs

题目

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?

解答

方法一：可用递归思想， 但太慢了，时间复杂度：O(2^lgn)，当n逐渐变大时，时间呈指数增长

int climbStairsRecur(int n){if(n==1) return 1;if(n==2) return 2;return climbStairsRecur(n-1)+climbStairsRecur(n-2);}

方法二：使用动态规划法填表，时间复杂度：O(n)

public class Solution {    public static int climbStairs(int n) {        int[] data=new int[n+1];    data[0]=0;    data[1]=1;    if(n>=2){    data[2]=2;    }    for(int i=3;i<=n;i++){    data[i]=data[i-1]+data[i-2];    }    return data[n];    }}

直接用变量存储

int climbStairs(int n){if(n<4) return n;int a=2,b=3,c=5;for(int i=5;i<=n;i++){a=c;    //此处a只是作为变量c=b+c;b=a;           }return c;}

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