diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index cd561ea6..4375b3f4 100644
--- a/notes/Leetcode 题解.md
+++ b/notes/Leetcode 题解.md
@@ -2727,30 +2727,63 @@ public int minPathSum(int[][] grid) {
题目描述：交易之后需要有一天的冷却时间。
- 
-
-```html
-s0[i] = max(s0[i - 1], s2[i - 1]); // Stay at s0, or rest from s2
-s1[i] = max(s1[i - 1], s0[i - 1] - prices[i]); // Stay at s1, or buy from s0
-s2[i] = s1[i - 1] + prices[i]; // Only one way from s1
-```
+ 
```java
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) return 0;
- int n = prices.length;
- int[] s0 = new int[n];
- int[] s1 = new int[n];
- int[] s2 = new int[n];
- s0[0] = 0;
- s1[0] = -prices[0];
- s2[0] = Integer.MIN_VALUE;
- for (int i = 1; i < n; i++) {
- s0[i] = Math.max(s0[i - 1], s2[i - 1]);
- s1[i] = Math.max(s1[i - 1], s0[i - 1] - prices[i]);
- s2[i] = Math.max(s2[i - 1], s1[i - 1] + prices[i]);
+ 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;
+ 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]);
}
- return Math.max(s0[n - 1], s2[n - 1]);
+ return Math.max(sell[N - 1], s2[N - 1]);
+}
+```
+
+**需要交易费用的股票交易**
+
+[Leetcode : 714. Best Time to Buy and Sell Stock with Transaction Fee (Medium)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/)
+
+```html
+Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
+Output: 8
+Explanation: The maximum profit can be achieved by:
+Buying at prices[0] = 1
+Selling at prices[3] = 8
+Buying at prices[4] = 4
+Selling at prices[5] = 9
+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。
+
+ 
+
+```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;
+ 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]);
+ }
+ return Math.max(sell[N - 1], s2[N - 1]);
}
```
@@ -3789,6 +3822,45 @@ public void moveZeroes(int[] nums) {
}
```
+**数组的度**
+
+[Leetcode : 697. Degree of an Array (Easy)](https://leetcode.com/problems/degree-of-an-array/description/)
+
+```html
+Input: [1,2,2,3,1,4,2]
+Output: 6
+```
+
+题目描述：数组的度定义为元素出现的最高频率，例如上面的数组度为 3。要求找到一个最小的子数组，这个子数组的度和原数组一样。
+
+```java
+public int findShortestSubArray(int[] nums) {
+ Map numsCnt = new HashMap<>();
+ Map numsLastIndex = new HashMap<>();
+ Map numsFirstIndex = new HashMap<>();
+ for (int i = 0; i < nums.length; i++) {
+ int num = nums[i];
+ numsCnt.put(num, numsCnt.getOrDefault(num, 0) + 1);
+ numsLastIndex.put(num, i);
+ if (!numsFirstIndex.containsKey(num)) {
+ numsFirstIndex.put(num, i);
+ }
+ }
+ int maxCnt = 0;
+ for (int num : nums) {
+ maxCnt = Math.max(maxCnt, numsCnt.get(num));
+ }
+ int ret = nums.length;
+ for (int i = 0; i < nums.length; i++) {
+ int num = nums[i];
+ int cnt = numsCnt.get(num);
+ if (cnt != maxCnt) continue;
+ ret = Math.min(ret, numsLastIndex.get(num) - numsFirstIndex.get(num) + 1);
+ }
+ return ret;
+}
+```
+
**对角元素相等的矩阵**
[Leetcode : 766. Toeplitz Matrix (Easy)](https://leetcode.com/problems/toeplitz-matrix/description/)
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