修正"需要冷却期的股票交易"算法，将意义不明的s1状态并入“buy”状态

原解答状态转移图列出四个状态，实际上三个就够了（将意义不明的s1状态并入“buy”状态）

Author: somone23412

notes/Leetcode 题解 - 动态规划.md

@@ -1055,27 +1055,28 @@ public int combinationSum4(int[] nums, int target) {
 
 题目描述：交易之后需要有一天的冷却时间。
 
-<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/ffd96b99-8009-487c-8e98-11c9d44ef14f.png" width="300px"> </div><br>
+
+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
+可用维特比算法求解
 
 ```java
 public int maxProfit(int[] prices) {
     if (prices == null || prices.length == 0) {
         return 0;
     }
     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];
+    int[] C = new int[N];
+    A[0] = 0;
+    B[0] = C[0] = -prices[0];
     for (int i = 1; i < N; i++) {
-        buy[i] = 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]) + prices[i];
-        s2[i] = Math.max(s2[i - 1], sell[i - 1]);
+        A[i] = Math.max(A[i - 1], C[i - 1]);
+        B[i] = Math.max(B[i - 1], A[i - 1] - prices[i]);
+        C[i] = B[i - 1] + prices[i];
     }
-    return Math.max(sell[N - 1], s2[N - 1]);
+    return Math.max(A[N - 1], C[N - 1]);
 }
 ```