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diff --git a/‎LeetCode/out/production/LeetCode/seventeen/Solution.class‎
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diff --git a/‎LeetCode/out/production/LeetCode/sixteen/Solution.class‎
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diff --git a/‎LeetCode/src/seventeen/Solution.java‎
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@@ -1,7 +1,6 @@
 # 加油站问题（gas-station）
 
 <center>知识点：贪心</center>
-
 ## 题目描述
 
 There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
@@ -18,14 +17,17 @@ The solution is guaranteed to be unique.
 如果从某个起点出发可以保证你开着你的车绕着赛道走一圈，则返回起点的加油站下标，否则返回-1
 返回起点加油站的下标。
 
-注意：给出的实例可以保证解决方案是唯一的。
+比如下图：
 
+![image-20190725222358833](http://picture-pool.oss-cn-beijing.aliyuncs.com/2019-07-25-142359.png)
 
+加油站上的数字代表该加油站具有的油量，圆圈上的数字代表该路段需要耗费的油量，那么小车该从哪个加油站出发才能保证顺利走一圈呢？
 
+注意：给出的实例可以保证解决方案是唯一的。
 
 ## 解题思路
 
-贪心的思路，计算从每一站出发，到达下一站后车里还能剩余的油，然后按照剩余油的多少进行排序，从剩余油最多的站出发，进行模拟。
+简单地想，我们当然是想从第1个加油站出发，到达下一个加油站时车里剩余的油尽可能多，这样车里剩下的油还可以继续后面的路程，所以，采取贪心的思路，首先计算从每一站出发，到达下一站后车里还能剩余的油，然后按照剩余油的多少进行排序，从可以剩余油最多的站出发，进行模拟行程，看最后可否走一圈，如果不可以，则换下一个起点。
 
 
 ## 代码
@@ -81,4 +83,8 @@ public int canCompleteCircuit(int[] gas, int[] cost) {
 
 ```
 
+可以求得上例中的答案为0，即从第0号加油站出发，就像下面这样：
+
+
+
 [完整代码](../src/seventeen/Solution.java)
@@ -94,8 +94,8 @@ int Partition(int[][] v, int low, int high) {
 
     public static void main(String[] args) {
         Solution solution = new Solution();
-        int[] gas = {2, 3, 1};
-        int[] cost = {3, 1, 2};
+        int[] gas = {3, 2, 5, 3, 4, 3};
+        int[] cost = {2, 3, 4, 4, 3, 4};
         int index = solution.canCompleteCircuit(gas, cost);
         System.out.println("startIndex is: " + index);
 
 
@@ -159,6 +159,7 @@
 - [**single-number-ii(出现一次的数2)**](./LeetCode/Doc/出现一次的数2.md)
 - [candy（糖果问题）](./LeetCode/Doc/糖果问题.md)
 - [Gas-station（加油站问题)](./LeetCode/Doc/加油站问题.md)
+- [clone-graph（图的复制）](./LeetCode/Doc/图的复制.md)