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topic
Suppose you are climbing stairs. It takes n steps to reach the top of the building.
You can climb 1 or 2 steps each time. How many different ways do you have to climb to the top of a building?
Note: Given n is a positive integer.
Example 1:
输入: 2
输出: 2
解释: 有两种方法可以爬到楼顶。
1. 1 阶 + 1 阶
2. 2 阶
Example 2:
输入: 3
输出: 3
解释: 有三种方法可以爬到楼顶。
1. 1 阶 + 1 阶 + 1 阶
2. 1 阶 + 2 阶
3. 2 阶 + 1 阶
Solution 1: Recursion
/**
* 思路:
* 通过观察爬楼梯符合斐波那契数
* 递归解决
*/
public int climbStairs(int n) {
if (n<=2)return n;
return climbStairs(n-1)+climbStairs(n-2);
}
Time complexity: On^2
Space complexity: O1
Solution 2: Memory array
/**
* 思路:
* 通过观察爬楼梯符合斐波那契数
* 递归解决
* 加入记忆数组,如果值在记忆数组中存在直接返回
*/
public int climbStairs(int n) {
int memory[]=new int[n+1];
return recursive(memory,n);
}
private int recursive(int[] memory, int n) {
if(n<=2)return n;
if (memory[n]!=0)return memory[n];
memory[n]=recursive(memory,n-1)+recursive(memory,n-2);
return memory[n];
}
Time complexity: On
Space complexity: On
Solution 3: Iteration (accumulation)
/**
* 思路:
* 通过观察爬楼梯符合斐波那契数
* 累加的求出第n层有多少策略
*/
public int climbStairs(int n) {
if(n<=2)return n;
int fist=1,second=2,result=0;
for (int i=3;i<=n;i++){
result=fist+second;
fist=second;
second=result;
}
return result;
}
Time complexity: On
Space complexity: O1
Solution 4: Iteration (array)
/**
* 思路:
* 通过观察爬楼梯符合斐波那契数
* 数组存放所有的fib数
*/
public int climbStairs(int n) {
if(n<=2)return n;
int result[]=new int[n];
result[0]=1;
result[1]=2;
for (int i=2;i<n;i++){
result[i]=result[i-1]+result[i-2];
}
return result[n-1];
}
Time complexity: On
Space complexity: On