今日记录
121. 买卖股票的最佳时机
class Solution {
public:
int maxProfit(vector<int>& prices) {
vector<vector<int>> dp(prices.size(), vector<int>(2, 0));
dp[0][0] = -prices[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
class Solution {
public:
int maxProfit(vector<int>& prices) {
vector<vector<int>> dp(prices.size(), vector<int>(2, 0));
dp[0][0] = -prices[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
class Solution {
public:
int maxProfit(vector<int>& prices) {
vector<vector<int>> dp(prices.size(), vector<int>(5, 0));
dp[0][1] = -prices[0];
dp[0][3] = -prices[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];
}
};
总结
要思考清楚dp数组的含义！！
