• ?123.買賣股票的最佳時機III??

class Solution {
public:int maxProfit(vector<int>& prices) {vector<vector<int>>dp(prices.size(),vector<int>(5));int len=prices.size();if(len==0)return 0;dp[0][0]=0;dp[0][1]=-prices[0];dp[0][2]=0;dp[0][3]=-prices[0];dp[0][4]=0;for(int i=1;i<len;i++){dp[i][0]=dp[i-1][0];dp[i][1]=max(dp[i-1][1],dp[i-1][0]-prices[i]);dp[i][2]=max(dp[i-1][2],dp[i-1][1]+prices[i]);dp[i][3]=max(dp[i-1][3],dp[i-1][2]-prices[i]);dp[i][4]=max(dp[i-1][4],dp[i-1][3]+prices[i]);}
return dp[prices.size()-1][4];}
};

• ?188.買賣股票的最佳時機IV?

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