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