力扣算法学习day35-1

力扣算法学习day35-1

188-买卖股票的最佳时机 IV

题目

代码实现

	class Solution {
    public int maxProfit(int k, int[] prices) {
        if(k == 0 || prices.length == 0){
            return 0;
        }
        // 原理和买卖股票的最佳时机 III 相同，从2笔交易拓展到k笔交易。
        // dp[i][0] = Math.max(dp[i-1][0],-prices[i])
        // dp[i][1] = Math.max(dp[i-1][1],prices[i] + dp[i-1][0])
        // dp[i][2] = Math.max(dp[i-1][2],dp[i-1][1] - prices[i])
        // dp[i][3] = Math.max(dp[i-1][3],dp[i-1][2] + prices[i])
        // dp[i][4] = Math.max(dp[i-1][4],dp[i-1][3] - prices[i])
        // dp[i][5] = Math.max(dp[i-1][5],dp[i-1][4] + prices[i])
        // ......
        int[][] dp = new int[prices.length][k * 2];

        // 初始化
        for(int i = 0;i < k * 2;i = i + 2){
            dp[0][i] = -prices[0];
            dp[0][i+1] = 0;
        }

        for(int i = 1;i < prices.length;i++){
            dp[i][0] = Math.max(dp[i-1][0],-prices[i]);
            // System.out.print(dp[i][0] + " ");
            dp[i][1] = Math.max(dp[i-1][1],prices[i] + dp[i-1][0]);
            // System.out.print(dp[i][1] + " ");
            for(int j = 2;j < k * 2;j = j + 2){
                dp[i][j] = Math.max(dp[i-1][j],dp[i-1][j-1] - prices[i]);
                // System.out.print(dp[i][j] + " ");
                dp[i][j+1] = Math.max(dp[i-1][j+1],dp[i-1][j] + prices[i]);
                // System.out.print(dp[i][j+1] + " ");
            }
            // System.out.println();
        }

        // 求最大值
        int max = dp[prices.length-1][1];
        for(int i = 3;i < k * 2;i = i + 2){
            if(dp[prices.length-1][i] > max){
                max = dp[prices.length-1][i];
            }
        }

        return max;
    }
}