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?
Note: Given n will be a positive integer.
Example 1:
Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps
Example 2:
Input: 3 Output: 3 Explanation: There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step
方法1:开始用的递归算法,由于效率不高,所以就换成了方法2 ,带记忆的动态规划
package leetcode;
import org.junit.Test;
/**
* @author zhangyu
* @version V1.0
* @ClassName: ClimbingStairs
* @Description: TOTO
* @date 2018/12/5 16:22
**/
public class ClimbingStairs {
@Test
public void fun() {
long startTime = System.currentTimeMillis();
int n = 44;
int num = climbStairs2(n);
long endTime = System.currentTimeMillis();
System.out.println(num);
System.out.println((endTime-startTime)/1000.0+"s");
}
private int climbStairs2(int n) {
if (n == 0) {
return 0;
} else if (n == 1) {
return 1;
} else if (n == 2) {
return 2;
} else {
return climbStairs2(n - 1) + climbStairs2(n - 2);
}
}
}
时间复杂度:O(n)
空间复杂度:O(n)
方法2:利用动态规划进行求解
package leetcode;
import org.junit.Test;
/**
* @author zhangyu
* @version V1.0
* @ClassName: ClimbingStairs
* @Description: TOTO
* @date 2018/12/5 16:22
**/
public class ClimbingStairs {
@Test
public void fun() {
long startTime = System.currentTimeMillis();
int n = 44;
int num = climbingStairs(n);
long endTime = System.currentTimeMillis();
System.out.println(num);
System.out.println((endTime-startTime)/1000.0+"s");
}
// 动态规划,记忆算法
private int climbingStairs(int n) {
if (n == 1) {
return 1;
}
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];
}
}
时间复杂度:O(n)
空间复杂度:O(n)