LeetCode 123.买卖股票的最佳时机III
题目链接:
LeetCode 123.买卖股票的最佳时机III
代码:
class Solution {
public:int maxProfit(vector<int>& prices) {int size = prices.size();if(size==0) return 0;//dp[i][0] 不操作//dp[i][1] 1买//dp[i][2] 1卖//dp[i][3] 2买//dp[i][4] 2卖vector<vector<int>>dp(size,vector<int>(5,0));dp[0][1]= -prices[0];dp[0][3] = -prices[0];for(int i=1;i<size;i++){dp[i][0] = 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[size-1][4];}
};
LeetCode 188.买卖股票的最佳时机IV
题目链接:
LeetCode 188.买卖股票的最佳时机IV
代码:
class Solution {
public:int maxProfit(int k, vector<int>& prices) {int size = prices.size();if(size==0) return 0;//dp[i][0] 不操作//dp[i][1] 1买//dp[i][2] 1卖//dp[i][3] 2买//dp[i][4] 2卖vector<vector<int>>dp(size,vector<int>(2*k+1,0));for(int i=1;i<2*k;i+=2){dp[0][i] = -prices[0];}for(int i=1;i<size;i++){for(int j=1;j<2*k+1;j = j+2){dp[i][j] = max(dp[i-1][j],dp[i-1][j-1]-prices[i]);dp[i][j+1] = max(dp[i-1][j+1],dp[i-1][j]+prices[i]);}}return dp[size-1][2*k];}
};