代码实现 public int fib(int n) {
if(n==0) return 0;
int[] dp = new int[n+1];
dp[0] = 0;
dp[1] = 1;
for (int i = 2; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
标签:契数,return,int,斐波,++,509,dp
来源: https://www.cnblogs.com/ywy1/p/15549647.html