题解 | 跳台阶
跳台阶
https://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
*
* @param number int整型
* @return int整型
*/
int jumpFloor(int number) {
if(number==1) return 1;
int a = 1, b = 1, res;
for(int i = 2;i<=number;i++){
res = a + b;
a = b;
b = res;
}
return res;
}
};
a为i-2,b为i-1,迭代O(1)空间复杂度计算结果
#牛客春招刷题训练营#https://www.nowcoder.com/discuss/727521113110073344
