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classSolution {
public:int mod = 1e9 + 7;
intnumWays(int n) {
if (n == 0)
return1;
if (n == 1)
return1;
int pre = 1;
int cur = 1;
int now;
for (int i = 2; i <= n; i++) {
now = (pre + now) % mod;
pre = cur;
cur = now;
}
return now;
}
};
The text was updated successfully, but these errors were encountered:
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changed the title
[剑指 Offer] 10- I. 青蛙跳台阶问题
[剑指 Offer] 10- II. 青蛙跳台阶问题
Jul 23, 2021
一只青蛙一次可以跳上1级台阶,也可以跳上2级台阶。求该青蛙跳上一个 n 级的台阶总共有多少种跳法。
答案需要取模 1e9+7(1000000007),如计算初始结果为:1000000008,请返回 1。
示例 1:
示例 2:
示例 3:
提示:
解法类似[剑指 Offer] 10- I. 斐波那契数列,只是初始值不同。代码如下:
The text was updated successfully, but these errors were encountered: