剑指 Offer第 8 天 动态规划（简单）

剑指 Offer 10- I. 斐波那契数列

题目思路

很明显是递推关系，直接写过去即可

题目代码

c++代码

class Solution {
    
    
private:
    int mod = 1e9+7;
    int f[110];
public:
    int fib(int n) {
    
    
        f[0]=0,f[1]=1;
        for(int i=2;i<=n;i++){
    
    
            f[i] = (f[i-1] + f[i-2])%mod;
        }
        return f[n];
    }
};

python代码

class Solution:
    def fib(self, n: int) -> int:
        a = 0
        b = 1
        c = 0
        mod = 1e9+7
        if not n :
            return a
        elif n==1 :
            return b
        else:
            for i in range(2,n+1):
                c = (a+b)%mod
                a = b%mod
                b = c%mod
            return int(c)

剑指 Offer 10- II. 青蛙跳台阶问题

题目思路

和上一个题目有异曲同工之妙。无非是n个台阶只能从第n-1个台阶和第n-2个台阶上来

题目代码

c++代码

class Solution {
    
    
private:
    int mod = 1e9+7;
    int f[110];
public:
    int numWays(int n) {
    
    
       f[0]=1,f[1]=1,f[2]=2;
        for(int i=2;i<=n;i++){
    
    
            f[i] = (f[i-1] + f[i-2])%mod;
        }
        return f[n];
    }
};

python代码

class Solution:
    def numWays(self, n: int) -> int:
        a = 1
        b = 1
        c = 0
        mod = 1e9+7
        if not n :
            return a
        elif n==1 :
            return b
        else:
            for i in range(2,n+1):
                c = (a+b)%mod
                a = b%mod
                b = c%mod
            return int(c)

剑指 Offer 63. 股票的最大利润

题目思路

典型题目，两个数组，第一组代表买下股票的最大盈利，第二组代表卖出后的最大盈利

题目代码

c++代码

class Solution {
    
    
public:
    int maxProfit(vector<int>& prices) {
    
    
        int s[100010],b[100010];
        int n = prices.size();
        for(int i=0;i<100010;i++){
    
    
            s[i]=-100000000;
            b[i]=0;
        }
        for(int i=1;i<=n;i++){
    
    
            s[i]=max(s[i-1],0-prices[i-1]);
            b[i]=max(b[i-1],s[i]+prices[i-1]);
            //cout<<s[i]<<" "<<b[i]<<endl;
        }
        return b[n];
    }
};

python代码

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        s = [-1000000]*100010
        b = [0]*100010
        n = len(prices)
        for i in range(1,n+1):
            s[i] = max(s[i-1],0-prices[i-1])
            b[i] = max(b[i-1],s[i]+prices[i-1])
        return b[n]

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