题目链接:力扣Fib数列
递归(超时)
class Solution { public int fib(int n) { if(n==0) return 0; if(n==1||n==2) return 1; else return fib(n-1)+fib(n-2); } }
递推
class Solution { public int fib(int n) { long f[] = new long[105]; f[1] = 1; f[2] = 1; for(int i = 3;i<=n;i++) { f[i] = (f[i-1]%(1000000007))+(f[i-2]%(1000000007)); } return (int)(f[n]%(1000000007)); } }