714. Best Time to Buy and Sell Stock with Transaction Fee

Your are given an array of integers 

prices

, for which the 

i

-th
element is the price of a given stock on day 

i

; and a non-negative integer 

fee

 representing
a transaction fee.

You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)

Return the maximum profit you can make.

Example 1:

Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
Output: 8
Explanation: The maximum profit can be achieved by:

Buying at prices[0] = 1Selling at prices[3] = 8Buying at prices[4] = 4Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Note:

0 < prices.length <= 50000

.

0 < prices[i] < 50000

.

0 <= fee < 50000

.

总感觉先求部分再合并的思路是不对的

以下是比赛是WA的code

package l714;

import java.util.ArrayLi
4000
st;
import java.util.List;

/**
* 总感觉先求部分再合并的思路是不对的
* @author lenovo
*
*/
class WA {
public int maxProfit(int[] prices, int fee) {
List<Integer> in = new ArrayList<Integer>(), out = new ArrayList<Integer>();
int ret = 0;
for(int i=0; i<prices.length; i++) {
int j = i+1, min = prices[i], minIdx = i;
for(; j<prices.length; j++) {
if(prices[j] < min) {
min = prices[j];
minIdx = j;
}
if(prices[j] - min > fee)	break;
}
if(j < prices.length) {
//        		ret += Math.max(prices[j] - min - fee, fee);
//        		l.add(prices[j] - min - fee);
in.add(minIdx);
out.add(j);
}
i = j - 1;
}

for(int i=0; i<in.size(); i++) {
if(i == in.size()-1)	ret += prices[out.get(i)]-prices[in.get(i)]-fee;
else {
//        		ret += Math.max(prices[out.get(i)]-prices[in.get(i)]-fee, b)
}
}

if(in.size() > 0)	ret=prices[out.get(0)]-prices[in.get(0)]-fee;
for(int i=1; i<in.size(); i++) {
ret += prices[out.get(i)]-prices[in.get(i)]-fee;
if(prices[out.get(i-1)] - prices[in.get(i)] < fee) {
ret += fee - prices[out.get(i-1)] + prices[in.get(i)];
}
}
//        	if(l.get(i) - fee > fee)	ret += l.get(i);
//        	else				ret += l.get(i) + fee;
//        	ret += l.get(i) + Math.max(0, fee-l.get(i));
return ret;
}
}

正确的思路是DP，找每个节点的状态：没有stock和有stock

class Solution {
public int maxProfit(int[] prices, int fee) {
int s0 = 0, s1 = Integer.MIN_VALUE;
for(int p : prices) {
int tmp = s0;
s0 = Math.max(s0, s1+p);
s1 = Math.max(s1, tmp-p-fee);
}
return s0;
}
}

或者状态复杂点

This problem is just like the other stock problems.
At given day, we can do 1 out of 4 things:

buy stock
hold stock
do nothing with empty portfolio
sell stock
We have 4 arrays with the length of # of the days, recording the max profit at given day if we do given operation.

Here is the code:

class Solution {
public int maxProfit(int[] prices, int fee) {
if(prices.length <= 1) return 0;
int[] buy = new int[prices.length];
int[] hold = new int[prices.length];
int[] skip = new int[prices.length];
int[] sell = new int[prices.length];
// the moment we buy a stock, our balance should decrease
buy[0] = 0 - prices[0];
// assume if we have stock in the first day, we are still in deficit
hold[0] = 0 - prices[0];
for(int i = 1; i < prices.length; i++){
// We can only buy on today if we sold stock
// or skipped with empty portfolio yesterday
buy[i] = Math.max(skip[i-1], sell[i-1]) - prices[i];
// Can only hold if we bought or already holding stock yesterday
hold[i] = Math.max(buy[i-1], hold[i-1]);
// Can skip only if we skipped, or sold stock yesterday
skip[i] = Math.max(skip[i-1], sell[i-1]);
// Can sell only if we bought, or held stock yesterday
sell[i] = Math.max(buy[i-1], hold[i-1]) + prices[i] - fee;
}
// Get the max of all the 4 actions on the last day.
int max = Math.max(buy[prices.length - 1], hold[prices.length - 1]);
max = Math.max(skip[prices.length - 1], max);
max = Math.max(sell[prices.length - 1], max);
return Math.max(max, 0);
}
}