auto commit

notes/Leetcode 题解.md

@@ -828,6 +828,28 @@ public int maxSubArray(int[] nums) {
 }
 ```
 
+**买入和售出股票最大的收益** 
+
+[121. Best Time to Buy and Sell Stock (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/)
+
+题目描述：只进行一次交易。
+
+只要记录前面的最小价格，将这个最小价格作为买入价格，然后将当前的价格作为售出价格，查看当前收益是不是最大收益。
+
+```java
+public int maxProfit(int[] prices) {
+    int n = prices.length;
+    if (n == 0) return 0;
+    int soFarMin = prices[0];
+    int max = 0;
+    for (int i = 1; i < n; i++) {
+        if (soFarMin > prices[i]) soFarMin = prices[i];
+        else max = Math.max(max, prices[i] - soFarMin);
+    }
+    return max;
+}
+```
+
 ## 二分查找
 
 **正常实现** 
@@ -3340,27 +3362,6 @@ public int maxProfit(int[] prices, int fee) {
 }
 ```
 
-**买入和售出股票最大的收益** 
-
-[121. Best Time to Buy and Sell Stock (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/)
-
-题目描述：只进行一次交易。
-
-只要记录前面的最小价格，将这个最小价格作为买入价格，然后将当前的价格作为售出价格，查看当前收益是不是最大收益。
-
-```java
-public int maxProfit(int[] prices) {
-    int n = prices.length;
-    if (n == 0) return 0;
-    int soFarMin = prices[0];
-    int max = 0;
-    for (int i = 1; i < n; i++) {
-        if (soFarMin > prices[i]) soFarMin = prices[i];
-        else max = Math.max(max, prices[i] - soFarMin);
-    }
-    return max;
-}
-```
 
 **只能进行两次的股票交易**