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                斐波那契數列的公式為 ``` f(0) = 0 f(1) = 1 f(n) = f(n-1) + f(n-2) ``` 最直觀的方式是使用遞歸,但是遞歸需要一定的棧空間,而且有重復計算,例如:f(8) = f(7) + f(6),f(7) = f(6) + f(5) 這樣,會多次計算,可以使用循環的方式,每次計算之后累加,如從f(2)開始,一直計算到f(n),期間一直保存這f(cur-1),f(cur-2) 遞歸程序 ``` public int get(int i) { if(i<0)return -1; if(i == 0||i == 1)return i; return get(i-1) + get(i-2); } ``` 循環程序 ``` public int get(int i) { if(i < 0) return -1; if(i == 0||i == 1)return i; int N$2 = 0, N$1 = 1,idx = 2,ret=0; while(idx <= i) { // 0 1 1 2 3 5 8 ret = N$2 + N$1; N$2 = N$1; N$1 = ret; idx++; } return ret; } ``` 假設你正在爬樓梯,需要n步你才能到達頂部。但每次你只能爬一步或者兩步,你能有多少種不同的方法爬到樓頂部? 比如n=3,1+1+1=1+2=2+1=3,共有3種不同的方法,返回3 **規則** 輸入: int n 樓梯的長度 輸出:int 多少種方法 case: ``` n 小于0 n等于0 n等于1 n等于2 ``` **思路** 假設第n個臺階有f(n)種方法,f(n-1)種再加上一步1個臺階就到達n了,此時有f(n-1)種,f(n-2)樓梯的時候再加上一步2個臺階就到達n了,此時有f(n-2)種,因此:f(n) = f(n-1)+f(n-2),f(0) = 0 ,f(1) = 1 f(2) = 2 **代碼** ``` public int climbStairs(int i) { if (i <= 0) { return 0; } if (i == 1) { return i; } int N$2 = 1, N$1 = 1, idx = 2, ret = 0; while (idx <= i) { // 0 1 1 2 3 5 8 ret = N$2 + N$1; N$2 = N$1; N$1 = ret; idx++; } return ret; } ```
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