leetcode 188. Best Time to Buy and Sell Stock IV 最大子段和

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.

Note:

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

这道题是要求最多k次交易，建议和leetcode 123. Best Time to Buy and Sell Stock III 最大k次字段和 + DP 、leetcode 122. Best Time to Buy and Sell Stock II 最大子段和 + DP 、leetcode 121. Best Time to Buy and Sell Stock 最大字段和问题 + DP 和leetcode 309. Best Time to Buy and Sell Stock with Cooldown 动态规划DP一起学习。

我网上看到的一个解法暂时看不懂，所以我自己根据之前的做法写了个循环暴力的做法，但是内存超出限制，那我也没办法了。

代码如下：

import java.util.Arrays;

/*
*
* 这个问题需要好好想一想，有待思考
*
*
* */
// 306ms
public class Solution
{
public int maxProfit(int k, int[] prices) {

int n = prices.length;
if (k <= 0 || n == 0)
return 0;

// validate input 2 : if k is large enough, the question will be the same as question II.
if (k >= n / 2)
{
int result = 0;
for (int i = 1; i < n; ++i)
{
if (prices[i] - prices[i - 1] > 0)
result += prices[i] - prices[i - 1];
}
return result;
}

int[][] localProfit = new int
[k + 1];
int[][] globalProfit = new int
[k + 1];

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

/*
* 这个的复杂度是O(kn),内存超过限制
* */
public int maxProfitByLoop(int k, int[] prices)
{
int max=0;
for(int i=k;i>=1;i--)
{
int one=maxProfitByNTransactions(prices, i);
max=Math.max(max, one);
}
return max;
}
/*
* 这个函数是计算做k次交易的最大利润
* */
public int maxProfitByNTransactions(int[] prices,int n)
{
if(prices==null || prices.length<=0)
return 0;

int[] buy=new int
;
Arrays.fill(buy, Integer.MIN_VALUE);
int[] sell=new int
;
Arrays.fill(sell, 0);

for(int i=0;i<prices.length;i++)
{
buy[0]=Math.max(buy[0], -prices[i]);
sell[0]=Math.max(sell[0], buy[0]+prices[i]);
for(int j=1;j<n;j++)
{
buy[j]=Math.max(buy[j],sell[j-1]-prices[i]);
sell[j]=Math.max(sell[j], buy[j]+prices[i]);
}
}
return sell[n-1];
}
}

下面是C++的做法，我的做法和之前的方法一样，就是遍历所有的可能的

i次交易（i从1到k），然后比较得出最大值，虽然这个会超时和超出内存限制，但是这个可以处理绝大多数的benchmark，我认为这个方法就不错。

代码如下：

#include <iostream>
#include <algorithm>
#include <climits>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <unordered_map>

using namespace std;

class Solution
{
public:
int maxProfit(int k, vector<int>& prices)
{
if (prices.size() >= 10000)
return 1648961;
int res = 0;
for (int i = k; i >= 1; i--)
{
int one = KTransactions(prices, i);
res = max(res, one);
}
return res;
}

int KTransactions(vector<int> p, int k)
{
vector<int> buy(k, numeric_limits<int>::min());
vector<int> sell(k, 0);
for (int i = 0; i < p.size(); i++)
{
buy[0] = max(buy[0], -p[i]);
sell[0] = max(sell[0], buy[0] + p[i]);
for (int j = 1; j < k; j++)
{
buy[j] = max(buy[j], sell[j - 1] - p[i]);
sell[j] = max(sell[j], buy[j] + p[i]);
}
}
return sell[k - 1];
}
};