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#yyds干货盘点# 面试必刷TOP101:跳台阶
source link: https://blog.51cto.com/u_15488507/5717160
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#yyds干货盘点# 面试必刷TOP101:跳台阶
精选 原创1.简述:
描述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(先后次序不同算不同的结果)。
数据范围:
要求:时间复杂度: ,空间复杂度:
示例1青蛙要跳上两级台阶有两种跳法,分别是:先跳一级,再跳一级或者直接跳两级。因此答案为2
2.代码实现:
public class Solution {
public int jumpFloor(int target) {
//从0开始,第0项是0,第一项是1
if(target <= 1)
return 1;
int res = 0;
int a = 0;
int b = 1;
//因n=2时也为1,初始化的时候把a=0,b=1
for(int i = 2; i <= target; i++){
//第三项开始是前两项的和,然后保留最新的两项,更新数据相加
res = (a + b);
a = b;
b = res;
}
return res;
}
}
public int jumpFloor(int target) {
//从0开始,第0项是0,第一项是1
if(target <= 1)
return 1;
int res = 0;
int a = 0;
int b = 1;
//因n=2时也为1,初始化的时候把a=0,b=1
for(int i = 2; i <= target; i++){
//第三项开始是前两项的和,然后保留最新的两项,更新数据相加
res = (a + b);
a = b;
b = res;
}
return res;
}
}
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