做公益网站需要什么资质竞价排名广告

 

目录

 

一、按摩师

二、打家劫舍

 三、删除并获得点数

 四、粉刷房子

 五、买卖股票的最佳时机

 六、买卖股票的最佳时机（含手续费）

七、买卖股票的最佳时机III

 八、买卖股票的最佳时机IV

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一、按摩师

class Solution {
public:int massage(vector<int>& nums) {int n =nums.size();if(n == 0) return 0;vector<int> f(n);auto g = f;f[0] = nums[0];for(int i = 1;i < n;i++){f[i] = g[i-1] + nums[i];g[i] = max(f[i-1],g[i-1]);}return max(f[n-1],g[n-1]);}
};

二、打家劫舍

class Solution {
public:int rob1(vector<int>& nums,int l,int r) {if(l>r) return 0;int n =nums.size();if(n == 0) return 0;vector<int> f(n);auto g = f;f[l] = nums[l];for(int i = l;i <= r;i++){f[i] = g[i-1] + nums[i];g[i] = max(f[i-1],g[i-1]);}return max(f[r],g[r]);}int rob(vector<int>& nums) {int n = nums.size();int ret1 = rob1(nums,2,n-2)+nums[0];int ret2 = rob1(nums,1,n-1);return max(ret1,ret2);}
};

 三、删除并获得点数

class Solution {
public:int deleteAndEarn(vector<int>& nums) {int n = nums.size();const int N = 10001;int arr[N] = {0}; for(auto e : nums){arr[e] += e;}vector<int> f(N);auto g = f;for(int i = 1;i < N;i++){f[i] = g[i-1] + arr[i];g[i] = max(f[i-1],g[i-1]);}return max(f[N-1],g[N-1]);}
};

 四、粉刷房子

 

class Solution {
public:int minCost(vector<vector<int>>& costs) {int n = costs.size();vector<vector<int>> dp(n+1,vector<int>(3));for(int i = 1;i <= n;i++){dp[i][0] = costs[i-1][0] + min(dp[i-1][1],dp[i-1][2]);dp[i][1] = costs[i-1][1] + min(dp[i-1][0],dp[i-1][2]);dp[i][2] = costs[i-1][2] + min(dp[i-1][0],dp[i-1][1]);}return min(dp[n][0],min(dp[n][1],dp[n][2]));}
};

 五、买卖股票的最佳时机

 

 

  

class Solution {
public:int maxProfit(vector<int>& prices) {int n = prices.size();vector<vector<int>> dp(n,vector<int>(3));dp[0][0] = -prices[0];for(int i = 1;i < n;i++){dp[i][0] = max(dp[i-1][0],dp[i-1][1] - prices[i]);dp[i][1] = max(dp[i-1][1],dp[i-1][2]);dp[i][2] = dp[i-1][0]+prices[i];}return max(dp[n-1][1],dp[n-1][2]);}
};

 六、买卖股票的最佳时机（含手续费）

上一题用的是二维数组的第二维来表示多种状态，是因为状态比较多，如果像此题只有两种状态，就可以用两个函数，本质上是一样的。 

 

class Solution {
public:int maxProfit(vector<int>& prices, int fee) {int n = prices.size();vector<int> f(n);auto g = f;f[0] = -prices[0];for(int i = 1;i < n;i++){f[i] = max(f[i-1],g[i-1] - prices[i]);g[i] = max(g[i-1],f[i-1]+prices[i]- fee);}return g[n-1];}
};

七、买卖股票的最佳时机III

 

 

class Solution {
public:const int INF = 0x3f3f3f3f;int maxProfit(vector<int>& prices) {int n = prices.size();vector<vector<int>> f(n,vector<int>(3,-INF));auto g = f;f[0][0] = -prices[0];g[0][0] = 0;for(int i = 1;i < n;i++){for(int j = 0;j < 3;j++){f[i][j] = max(f[i-1][j],g[i-1][j] - prices[i]);g[i][j] = g[i-1][j];if(j >= 1)g[i][j] = max(g[i-1][j],f[i-1][j-1]+prices[i]);}}   int ret = 0;for(int i = 0;i < 3;i++){ret = max(ret,g[n-1][i]);}return ret;}
};

 八、买卖股票的最佳时机IV

 

class Solution {
public:int maxProfit(int k, vector<int>& prices) {int  n = prices.size();k = min(k,n/2); const int INF = 0x3f3f3f3f;vector<vector<int>> f(n,vector<int>(k+1,-INF));//注意是k+1auto g = f;f[0][0] = -prices[0];g[0][0] = 0;for(int i = 1;i < n;i++){for(int j = 0;j <= k;j++){f[i][j] = max(f[i-1][j],g[i-1][j]-prices[i]);g[i][j] = g[i-1][j];if(j >= 1)g[i][j] = max(g[i-1][j],f[i-1][j-1]+prices[i]);}}int ret = 0;for(int j = 0;j <= k;j++){ret = max(ret,g[n-1][j]);}return ret;}
};