本文共 1995 字，大约阅读时间需要 6 分钟。

题目

解析

class Solution {
   public:    int maxProfit(int k, vector
   
    & prices) {
           if (prices.empty()) {
               return 0;        }        int n = prices.size();        k = min(k, n / 2);        vector
    
     
      > buy(n, vector
      
       (k + 1));        vector
       
        
         > sell(n, vector
         
          (k + 1)); buy[0][0] = -prices[0]; sell[0][0] = 0; for (int i = 1; i <= k; ++i) { buy[0][i] = sell[0][i] = INT_MIN / 2; } for (int i = 1; i < n; ++i) { buy[i][0] = max(buy[i - 1][0], sell[i - 1][0] - prices[i]); for (int j = 1; j <= k; ++j) { buy[i][j] = max(buy[i - 1][j], sell[i - 1][j] - prices[i]); sell[i][j] = max(sell[i - 1][j], buy[i - 1][j - 1] + prices[i]); } } return *max_element(sell[n - 1].begin(), sell[n - 1].end()); }};/*作者：LeetCode-Solution链接：https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iv/solution/mai-mai-gu-piao-de-zui-jia-shi-ji-iv-by-8xtkp/来源：力扣（LeetCode）著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。*/
         
        
       
      
     
    
   

class Solution {
   public:    int maxProfit(int k, vector
   
    & prices) {
           if (prices.empty()) {
               return 0;        }        int n = prices.size();        k = min(k, n / 2);        vector
    
      buy(k + 1);        vector
     
       sell(k + 1);        buy[0] = -prices[0];        sell[0] = 0;        for (int i = 1; i <= k; ++i) {
               buy[i] = sell[i] = INT_MIN / 2;        }        for (int i = 1; i < n; ++i) {
               buy[0] = max(buy[0], sell[0] - prices[i]);            for (int j = 1; j <= k; ++j) {
                   buy[j] = max(buy[j], sell[j] - prices[i]);                sell[j] = max(sell[j], buy[j - 1] + prices[i]);               }        }        return *max_element(sell.begin(), sell.end());    }};/*作者：LeetCode-Solution链接：https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iv/solution/mai-mai-gu-piao-de-zui-jia-shi-ji-iv-by-8xtkp/来源：力扣（LeetCode）著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。*/