LeetCode-188.Best Time to Buy and Sell Stock IV

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most ktransactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Example 1:

Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
             Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
 
使用动态规划

  public int maxProfit(int k, int[] prices) {//dp mytip
         if(null==prices||0==prices.length||0>=k){
             return 0;
         }
         int max =0;
         if(k>prices.length){//若k过大 优化
             for (int i = 0; i < prices.length-1; i++) {
                 if(prices[i]<prices[i+1]){
                     max+= prices[i+1]-prices[i];
                 }
             }
         }
         else{
             int[][][] states = new int[prices.length][k+1][2];//状态表示第i个数在第j此交易中，有无股票时（0为无，1为有）的利益；//因为只保存上一个数时的利益，所以states可优化为[k+1][2]
             for(int i=0;i<=k;i++){//初始化第1个数的状态
                 states[0][i][1]=-prices[0];
             }
             for(int i=1;i<prices.length;i++){
                 for(int j=0;j<=k;j++){
                     if(j==0){
                         states[i][j][0] = states[i-1][j][0];//防止j-1溢出
                     }
                     else{
                         states[i][j][0] = Math.max(states[i-1][j][0],states[i-1][j-1][1]+prices[i]);
                     }
                     states[i][j][1] = Math.max(states[i-1][j][1],states[i-1][j][0]-prices[i]);
                 }
             }
             for(int i=0;i<=k;i++){
                 max = max>states[prices.length-1][i][0]?max:states[prices.length-1][i][0];
             }
         }
 
         return max;
     }

优化空间

public int maxProfit(int k, int[] prices) {//dp my
        if(null==prices||0==prices.length||0>=k){
            return 0;
        }
        int max =0;
        if(k>prices.length){
            for (int i = 0; i < prices.length-1; i++) {
                if(prices[i]<prices[i+1]){
                    max+= prices[i+1]-prices[i];
                }
            }
        }
        else{
            int[][] states = new int[k+1][2];
            states[0][1] = -prices[0];
            for(int i=0;i<=k;i++){
                states[i][1]=-prices[0];
            }
            for(int i=1;i<prices.length;i++){
                for(int j=0;j<=k;j++){
                    states[j][1] = Math.max(states[j][1],states[j][0]-prices[i]);
                    if(j==0){
                        states[j][0] = states[j][0];
                    }
                    else{
                        states[j][0] = Math.max(states[j][0],states[j-1][1]+prices[i]);
                    }
 
                }
            }
            for(int i=0;i<=k;i++){
                max = max>states[i][0]?max:states[i][0];
            }
        }
 
        return max;
    }

相关题

买卖股票的最佳时间1 LeetCode121 https://www.cnblogs.com/zhacai/p/10429264.html

买卖股票的最佳时间2 LeetCode122 https://www.cnblogs.com/zhacai/p/10596627.html

买卖股票的最佳时间3 LeetCode123 https://www.cnblogs.com/zhacai/p/10645571.html

买卖股票的最佳时间冷冻期 LeetCode309 https://www.cnblogs.com/zhacai/p/10655970.html

买卖股票的最佳时间交易费 LeetCode714 https://www.cnblogs.com/zhacai/p/10659288.html