10- I、斐波那契数列（fib）

Offer 10- I. 斐波那契数列（fib）

1. python

class Solution:
    def fib(self, n: int) -> int:
        a,b = 0,1
        for _ in range(n):
            a,b = b,a+b
        return a % 1000000007
        

class Solution:
    def fib(self, n: int) -> int:
        if 0==n:
            return 0
        if 1==n:
            return 1;
        dp = [0]*(n+1)
        dp[0]=0
        dp[1]=1
        for i in range(2,n+1):
            dp[i]=dp[i-1]+dp[i-2]
        return dp[n]%1000000007

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2. Java

class Solution {
    
    
    public int fib(int n) {
    
    
        if(0==n){
    
    
            return 0;
        }
        if(1==n){
    
    
            return 1;
        }
        int[] dp = new int[n+1];
        dp[0]=0;
        dp[1]=1;
        int i=2;
        while (i<=n){
    
    
            dp[i]=dp[i-1]+dp[i-2];
            dp[i]%=1000000007;
            i++;
        }
        return dp[n];

    }
}

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3. C++

class Solution {
public:
    int fib(int n) {
        if(n==0){
            return 0;
        }
        if(n==1){
            return 1;
        }
        vector<int> dp(n+1);
        dp[0]=0;
        dp[1]=1;
        for(int i=2;i<=n;i++){
            dp[i]=dp[i-1]+dp[i-2];
            dp[i]=dp[i]%1000000007;
        }
        return dp[n];
    }
};

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