Best Time to Buy and Sell Stock IV
描述
Say you have an array for which the i-th 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.
Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
分析
设两个状态，global[i][j] 表示i天前最多可以进行j次交易的最大利润，local[i][j]表示i天前最多可以进行j次交易，且在第i天进行了第j次交易的最大利润。状态转移方程如下：
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])
关于global的状态转移方程比较简单，不断地和已经计算出的local进行比较，把大的保存在global中。
关于local的状态转移方程，取下面二者中较大的一个：
- 全局前
i-1天进行了j-1次交易，然后然后加上今天的交易产生的利润（如果赚钱就交易，不赚钱就不交易，什么也不发生，利润是0） - 局部前
i-1天进行了j次交易，然后加上今天的差价（local[i-1][j]是第i-1天卖出的交易，它加上diff后变成第i天卖出，并不会增加交易次数。无论diff是正还是负都要加上，否则就不满足local[i][j]必须在最后一天卖出的条件了）
注意，当k大于数组的大小时，上述算法将变得低效，此时可以改为不限交易次数的方式解决，即等价于 "Best Time to Buy and Sell Stock II"。
解法1
// Best Time to Buy and Sell Stock IV
// Time Complexity: O(nk), Space Complexity: O(nk)
public class Solution {
public int maxProfit(final int k, final int[] prices) {
if (prices.length < 2 || k < 1) return 0;
if (k >= prices.length) return maxProfit(prices);
final int[][] local = new int[prices.length][k + 1];
final int[][] global = new int[prices.length][k + 1];
for (int i = 1; i < prices.length; i++) {
final int diff = prices[i] - prices[i - 1];
for (int j = 1; j < k+1; j++) {
local[i][j] = Math.max(
global[i - 1][j - 1] + Math.max(diff, 0),
local[i - 1][j] + diff);
global[i][j] = Math.max(global[i - 1][j], local[i][j]);
}
}
return global[prices.length - 1][k];
}
// Best Time to Buy and Sell Stock II
public static int maxProfit(final int[] prices) {
int sum = 0;
for (int i = 1; i < prices.length; i++) {
int diff = prices[i] - prices[i - 1];
if (diff > 0) sum += diff;
}
return sum;
}
}