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简化”需要交易费用的股票交易“题解，删除不必要的s状态

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原图画了四个状态实在没有必要，只需要两个状态
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notes/Leetcode 题解 - 动态规划.md
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@ -1057,7 +1057,7 @@ public int combinationSum4(int[] nums, int target) {
该题为马尔可夫过程，分为A观望，B持股，C冷却三个状态
状态转移图：A-(观望)->A, A-(买入)->B, B-(观望)->B, B-(卖出)->C, C-(冷却)->A
状态转移图：A-(观望)->A, A-(买入｜-price)->B, B-(观望)->B, B-(卖出|+price)->C, C-(冷却)->A
可用维特比算法求解
```java
@ -1099,24 +1099,22 @@ The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
题目描述：每交易一次，都要支付一定的费用。
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/1e2c588c-72b7-445e-aacb-d55dc8a88c29.png" width="300px"> </div><br>
分为A观望，B持股，两个状态
状态转移图：A-(观望)->A, A-(买入|-price)->B, B-(观望)->B, B-(卖出|+price|-fee)->A
```java
public int maxProfit(int[] prices, int fee) {
int N = prices.length;
int[] buy = new int[N];
int[] s1 = new int[N];
int[] sell = new int[N];
int[] s2 = new int[N];
s1[0] = buy[0] = -prices[0];
sell[0] = s2[0] = 0;
int[] A = new int[N];
int[] B = new int[N];
A[0] = 0;
B[0] = -prices[0];
for (int i = 1; i < N; i++) {
buy[i] = Math.max(sell[i - 1], s2[i - 1]) - prices[i];
s1[i] = Math.max(buy[i - 1], s1[i - 1]);
sell[i] = Math.max(buy[i - 1], s1[i - 1]) - fee + prices[i];
s2[i] = Math.max(s2[i - 1], sell[i - 1]);
A[i] = Math.max(A[i - 1], B[i - 1] + prices[i] -fee);
B[i] = Math.max(A[i - 1] - prices[i], B[i - 1]);
}
return Math.max(sell[N - 1], s2[N - 1]);
return A[N - 1];
}
```