Best Time to Buy and Sell Stock IV 解答

Question

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 k transactions.

For example, given prices = [4,4,6,1,1,4,2,5], and k = 2, return 6.

Answer

用DP解答。

local[i][j] 表示0～i的数，最多j次交易，并且最后一次交易必须包含prices[j]（即最后一天必须卖出），收益最大值。

global[i][j] 表示0~i的数，最多j次交易，收益最大值。

diff = prices[i] - prices[i-1] local[i][j] = max(global[i-1][j-1] + max(diff,0),local[i-1][j]+diff) global[i][j] = max(local[i][j], global[i-1][j])

local[i-1][j] + diff 表示第i-1天，买进prices[i-1]，第i天卖出。

由于local[i-1][j]表示最后一次交易必须包含prices[i-1]，即prices[i-1]必须卖出。所以可以把第i-1天的交易和第i天的交易合并，即成了最多j次交易，最后一天交易必须卖出prices[i]。 global[i-1][j-1] 表示前i-1天，最多j-1次交易。最后一天比较diff，如果diff>0，则第i-1天买进，第i天卖出；如果diff<0，则第i天买进，第i天卖出。

class Solution {
    /**
     * @param k: An integer
     * @param prices: Given an integer array
     * @return: Maximum profit
     */
    public int maxProfit(int k, int[] prices) {
        // write your code here
        if (k == 0) {
            return 0;
        }
        if (k >= prices.length / 2) {
            int profit = 0;
            for (int i = 1; i < prices.length; i++) {
                if (prices[i] > prices[i - 1]) {
                    profit += prices[i] - prices[i - 1];
                }
            }
            return profit;
        }
        int n = prices.length;
        int[][] mustsell = new int[n + 1][n + 1];   // mustSell[i][j] 表示前i天，至多进行j次交易，第i天必须sell的最大获益
        int[][] globalbest = new int[n + 1][n + 1];  // globalbest[i][j] 表示前i天，至多进行j次交易，第i天可以不sell的最大获益

        mustsell[0][0] = globalbest[0][0] = 0;
        for (int i = 1; i <= k; i++) {
            mustsell[0][i] = globalbest[0][i] = 0;
        }

        for (int i = 1; i < n; i++) {
            int gainorlose = prices[i] - prices[i - 1];
            mustsell[i][0] = 0;
            for (int j = 1; j <= k; j++) {
                mustsell[i][j] = Math.max(globalbest[(i - 1)][j - 1] + gainorlose,
                                            mustsell[(i - 1)][j] + gainorlose);
                globalbest[i][j] = Math.max(globalbest[(i - 1)][j], mustsell[i ][j]);
            }
        }
        return globalbest[(n - 1)][k];
    }
};