Day.35 | 121.买卖股票的最佳时机 122.买卖股票的最佳时机II 123.买卖股票的最佳时机III
121.买卖股票的最佳时机
要点：dp[i][0] 是第 i 天持有股票时的资产，dp[i][1] 是第 i 天卖出股票时的资产，只跟前一天有关
class Solution {
public:int maxProfit(vector<int>& prices) {if (prices.size()==1) return 0;vector<vector<int>> dp(prices.size(), vector<int>{0, 0});dp[0][0] = -prices[0];dp[0][1] = 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][1], prices[i] + dp[i - 1][0]);}return dp[prices.size() - 1][1];}
};122.买卖股票的最佳时机II
要点：没理解啊没理解
class Solution {
public:int maxProfit(vector<int>& prices) {vector<vector<int>> dp(prices.size(), vector<int>(2, 0));dp[0][0] = -prices[0];dp[0][1] = 0;for (int i = 1; i < prices.size(); 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][0] + prices[i]);}return dp[prices.size() - 1][1];}
};123.买卖股票的最佳时机III
要点：当天的状态会延续前一天的状态
class Solution {
public:int maxProfit(vector<int>& prices) {vector<vector<int>> dp(prices.size(), vector<int>(5, 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 < prices.size(); ++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];}
};