大家都知道斐波那契數列,現在要求輸入一個整數n,請你輸出斐波那契數列的第n項(從0開始,第0項爲0)。斐波那契數列屬於經典的遞歸問題,對於這題的求解,我們首先要知道斐波那契數列的狀態轉移式,即f[n]=f[n-1]+f[n-2]
,且在n=1或2時,f[n]=1。
public static int feibolaqiByCircle(int n) {
if (n < 0) {
return -1;
}
if (n <= 1) {
return n;
}
int sum = 1;// f(n-1)
int pre = 0;// f(n-2)
for (int i = 2; i <= n; i++) {
sum = sum + pre;
pre = sum - pre;
}
return 0;
}
public static int feibolaqiByRecursion(int n) {
if (n < 0) {
return -1;
}
if (n < 1) {
return n;
}
return feibolaqiByCircle(n - 1) + feibolaqiByCircle(n - 2);
}