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#yyds干货盘点# 解决剑指offer：跳台阶扩展问题
source link: https://blog.51cto.com/u_15488507/5372403
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#yyds干货盘点# 解决剑指offer：跳台阶扩展问题
原创1.简述：
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(n为正整数)总共有多少种跳法。
数据范围：
进阶：空间复杂度 ， 时间复杂度
2.代码实现：
public class Solution {
public int jumpFloorII(int target) {
int[] dp = new int[target + 1];
//初始化前面两个
dp[0] = 1;
dp[1] = 1;
//依次乘2
for(int i = 2; i <= target; i++)
dp[i] = 2 * dp[i - 1];
return dp[target];
}
}
public int jumpFloorII(int target) {
int[] dp = new int[target + 1];
//初始化前面两个
dp[0] = 1;
dp[1] = 1;
//依次乘2
for(int i = 2; i <= target; i++)
dp[i] = 2 * dp[i - 1];
return dp[target];
}
}
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