LintCode 查找斐波纳契数列中第 N 个数

0, 1, 1, 2, 3, 5, 8, 13, 21, 34 ...

1

0

2

1

10

34

题目就是斐波那契数算法的变形。关于斐波那契数，首先想到递归

class Solution{
public:
/**
* @param n: an integer
* @return an integer f(n)
*/
int fibonacci(int n) {
// write your code here
if ( n == 1 ) {
return 0;
} else if ( n == 2 || n == 3 ) {
return 1;
}
/*else {
return fibonacci(n-1) + fibonacci(n-2);
}
｝；

但是递归太耗费时间，结果超时。

再换成非递归形式

<pre name="code" class="cpp">class Solution{
public:
/**
* @param n: an integer
* @return an integer f(n)
*/
int fibonacci(int n) {
// write your code here
if ( n == 1 ) {
return 0;
} else if ( n == 2 || n == 3 ) {
return 1;
}
int s1 = 1;
int s2 = 1;
int i = 4;
int sum = 0;
while (i <= n) {
sum = s1 + s2;
s1 = s2;
s2 = sum;
i++;
}
return sum;
}
};