1137. 第 N 个泰波那契数

class Solution {
   public int tribonacci(int n) {
		  int T0 = 0;
		  int T1 = 1;
		  int T2 = 1;
		  int arr[] = new int [40];
		  arr[0] = 0;
		  arr[1] = 1;
		  arr[2] = 1;
		  for(int i = 3 ; i <= n ; i++) {
			  arr[i] = arr[i-2]+ arr[i-1]+arr[i-3];
		  }
	      return arr[n]; 	
	    }
	}

有一种更省空间 不用开一个array

class Solution {
    public int tribonacci(int n) {
        if (n==0)
			return 0;
		int T1 = 0;
		int T2 = 1;
		int T3 = 1;
		for (int i = 4; i <= n+1; i++) {
			int T4 = T1 + T2 + T3;
			T1 = T2;
			T2 = T3;
			T3 = T4;
		}
		return T3;
    }
}
原文地址:https://www.cnblogs.com/cznczai/p/11289364.html