算法 跳台阶

跳台阶

 

一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法。

 

解法一：

public class Solution {
    public int JumpFloor(int target) { // 递归
        if (target == 1) {
            return 1;
        }
        if (target == 2) {
            return 2;
        }
        return JumpFloor(target - 1) + JumpFloor(target - 2);
    }
}

 

解法二：

public class Solution {
    public int JumpFloor(int target) { // 非递归
        if (target == 1) {
            return 1;
        }
        if (target == 2) {
            return 2;
        }
        int a = 1, b = 2, c = 0;
        for (int i = 3; i <= target; i++) {
            c = a + b;
            a = b;
            b = c;
        }
        return c;
    }
}

 

 

扩展：一只青蛙一次可以跳上1级台阶，也可以跳上2级……它也可以跳上n级。求该青蛙跳上一个n级的台阶总共有多少种跳法。

解法一：

public class Solution {
    public int JumpFloorII(int target) {
       if (target == 1) {
            return 1;
        }
        if (target == 2) {
            return 2;
        }
        int n = 1;
        for (int i = 1; i < target; i++) { // 递归
            n += JumpFloorII(target - i);
        }
        return n;
    }
     
    public int sum(int[] array, int start, int end) {
        int sum = 0;
        for (int i = start; i < end; i++) {
            sum += array[i];
        }
        return sum;
    }
}

 解法二：

public class Solution {
    public int JumpFloorII(int target) { // 动态规划
       if (target == 1) {
            return 1;
        }
        if (target == 2) {
            return 2;
        }
        int[] array = new int[target];
        array[0] = 1;
        array[1] = 2;
        for (int i = 3; i <= target; i++) {
            array[i - 1] = sum(array, 0, i - 1) + 1;
        }
        return array[target - 1];
    }
     
    public int sum(int[] array, int start, int end) {
        int sum = 0;
        for (int i = start; i < end; i++) {
            sum += array[i];
        }
        return sum;
    }
}