一. 题目描述
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.
【数据结构与算法|leetcode笔记（Best Time to Buy and Sell Stock IV）】Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
二. 题目分析
这一题的难度要远高于前面几题，需要用到动态规划，代码参考了博客：http://www.cnblogs.com/grandyang/p/4295761.html
这里需要两个递推公式来分别更新两个变量local和global，然后求至少k次交易的最大利润。我们定义local[i][j]为在到达第i天时最多可进行j次交易并且最后一次交易在最后一天卖出的最大利润，此为局部最优。然后我们定义global[i][j]为在到达第i天时最多可进行j次交易的最大利润，此为全局最优。它们的递推式为：
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])
三. 示例代码
#include
#include
#include
#include
#include
using namespace std;
class Solution {
public:
int maxProfit(int k, vector &prices) {
if(prices.empty() || k == 0)
return 0;
if(k >= prices.size())
return solveMaxProfit(prices);
vector global(k + 1, 0);
vector local(k + 1, 0);
for(int i = 1;
i < prices.size();
i++) {
int diff = prices[i] - prices[i - 1];
for(int j = k;
j >= 1;
j--) {
local[j] = max(local[j] + diff, global[j - 1] + max(diff, 0));
global[j] = max(global[j], local[j]);
}
}return global[k];
}private:
int solveMaxProfit(vector &prices) {
int res = 0;
for(int i = 1;
i < prices.size();
i++) {
int diff = prices[i] - prices[i - 1];
if(diff > 0)
res += diff;
}
return res;
}
};
四. 小结
参考链接：http://www.cnblogs.com/grandyang/p/4295761.html
