feat: add 121

Author: Blankj

README.md

@@ -38,6 +38,7 @@
 |112|[Path Sum][112]|Tree, Depth-first Search|
 |118|[Pascal's Triangle][118]|Array|
 |119|[Pascal's Triangle II][119]|Array|
+|121|[Best Time to Buy and Sell Stock][121]|Array, Dynamic Programmin|
 
 
 ## Medium
@@ -91,6 +92,7 @@
 [112]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/112/README.md
 [118]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/118/README.md
 [119]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/119/README.md
+[121]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/121/README.md
 
 [008]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/008/README.md
 [019]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md

note/119/README.md

@@ -2,7 +2,7 @@
 
 ## Description
 
-Given an index *k*, return the *k*th row of the Pascal's triangle.
+Given an index *k*, return the *k*<sup>th</sup> row of the Pascal's triangle.
 
 For example, given *k* = 3,
 Return `[1,3,3,1]`.

note/121/README.md

@@ -0,0 +1,57 @@
+# [Best Time to Buy and Sell Stock][title]
+
+## Description
+
+Say you have an array for which the *i*<sup>th</sup> element is the price of a given stock on day *i*.
+
+If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
+
+**Example 1:**
+
+```
+Input: [7, 1, 5, 3, 6, 4]
+Output: 5
+
+max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
+
+```
+
+**Example 2:**
+
+```
+Input: [7, 6, 4, 3, 1]
+Output: 0
+
+In this case, no transaction is done, i.e. max profit = 0.
+```
+
+**Tags:** Array, Dynamic Programmin
+
+
+## 思路
+
+题意是给出一个数组代表每天的股票金额，让你在最多买卖一次的情况下算出最大的收益额，最简单的就是模拟即可，每次记录当前值减去最小值的差值，与上一次的进行比较然后更新最大值即可。
+
+``` java
+class Solution {
+ public int maxProfit(int[] prices) {
+ int max = 0, minPrice = Integer.MAX_VALUE;
+ for (int i = 0; i < prices.length; ++i) {
+ if (prices[i] < minPrice) minPrice = prices[i];
+ int delta = prices[i] - minPrice;
+ if (delta > max) max = delta;
+ }
+ return max;
+ }
+}
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode.com/problems/best-time-to-buy-and-sell-stock
+[ajl]: https://github.com/Blankj/awesome-java-leetcode