Leetcode DAY 49~50：买卖股票的最佳时机 1 2 3 4

class Solution {
public:int maxProfit(vector<int>& prices) {//贪心int low = INT_MAX;int res = 0;for(int i = 0; i < prices.size(); i++) {low = min(low, prices[i]); //左最小价格res = max(res, prices[i] - low); //当前-左最小的max}return res;}
};

2、动态规划

dp[i][0]表示第i天持有股票最大现金 -> max(前一天持有股票的最大现金, 前一天买入股票的现金)
dp[i][1]表示第i天不持有股票最大现金 -> max(前一天不持有股票最大现金, 前一天持有第i天买入的现金)

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], -prices[i]);dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i]);}return dp[prices.size() - 1][1];}
};

122.买卖股票的最佳时机II

1、dp

dp[i][0]表示第i天持有股票时的所的现金

dp[i][1]表示第i天不持有股票所得的现金

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])

class Solution {
public:int maxProfit(vector<int>& prices) {int n = prices.size();vector<vector<int>> dp(n, vector<int>(2,0));dp[0][0] = -prices[0];dp[0][1] = 0;for(int i = 1; i < n; 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[n - 1][1];}
};

2、其他

class Solution:def maxProfit(self, prices: List[int]) -> int:res = 0for i in range(1, len(prices)):if prices[i] - prices[i - 1] > 0:res += prices[i] - prices[i - 1]return res

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

class Solution {
public:int maxProfit(int k, vector<int>& prices) {int n = prices.size();vector<vector<int>> dp(n, 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 < n; i++) {for(int j = 1; j <= 2 * k; j++) {dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + (1 - 2 * (j % 2)) * prices[i]);//1 - 2 * (j % 2)用来区分 奇数和偶数}}return dp[n - 1][2 *k];}
};