fix formula img show

README.md

@@ -247,7 +247,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 337 |[House Robber III](https://leetcode.com/problems/house-robber-iii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/337.%20House%20Robber%20III.md)|Medium|  |
 | 338 |[Counting Bits](https://leetcode.com/problems/counting-bits/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/338.%20Counting%20Bits.md)|Medium|  |
 | 341 |[Flatten Nested List Iterator](https://leetcode.com/problems/flatten-nested-list-iterator/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/341.%20Flatten%20Nested%20List%20Iterator.md)|Medium|  |
-| 342 |[Power of Four](https://leetcode.com/problems/power-of-four)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/342.%20Power%20of%20Four.md)|Easy|  |
+| 342 |[Power of Four](https://leetcode.com/problems/power-of-four)|[C++](solutions/342.%20Power%20of%20Four.md)|Easy|  |
 | 343 |[Integer Break](https://leetcode.com/problems/integer-break/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/343.%20Integer%20Break.md)|Medium|  |
 | 344 |[Reverse String](https://leetcode.com/problems/reverse-string)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/344.%20Reverse%20String.md)|Easy|  |
 | 345 |[Reverse Vowels of a String](https://leetcode.com/problems/reverse-vowels-of-a-string)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/345.%20Reverse%20Vowels%20of%20a%20String.md)|Easy|  |

solutions/342. Power of Four.md

@@ -14,15 +14,20 @@
 
 第1个条件等价于去掉最后的那个1后整个数变为0，即 `num & (num - 1) == 0`；(**注意学习这种去掉二进制最后一个1的方法**)   
 第2个条件等价于 `(num | mask) == mask`, 其中`mask = 0b01010101010101010101010101010101`。
-## 思路二
-还是基于思路一的两个条件，我们知道如果只满足思路一的条件1的数可能是2^n也可能是4^n, 怎样排除掉2^n呢？   
-我们知道：   
-<img src="https://latex.codecogs.com/svg.latex?\Large&space;&&2^n=(3-1)^n=C_n^03^0(-1)^n+C_{n-1}^13^1(-1)^{n-1}+......+C_n^n3^n(-1)^0&&" />   
-所以, 
-* 1.n为偶数时既是2的幂也是4的幂，此时(-1)^n==1所以(2^n-1）% 3==0;
-* 2.n为奇数是只是2的幂但不是4的幂, 此时(-1)^n=-1所以(2^n-1）% 3==1；
 
-故可以用(2^n-1）% 3是否等于0来等价判断思路一的条件2, 将2的幂和4的幂区分开。
+## 思路二
+还是基于思路一的两个条件，我们知道如果只满足思路一的条件1的数可能是`2^n`也可能是`4^n`, 怎样排除掉`2^n`呢？   
+我们知道：   
+<div align=center>
+<img width="500" src="img/342.png"/>
+</div>
+
+所以, 
+
+* 1.n为偶数时既是2的幂也是4的幂，此时`(-1)^n==1`所以`(2^n - 1）% 3==0`;
+* 2.n为奇数是只是2的幂但不是4的幂, 此时`(-1)^n=-1`所以`(2^n - 1）% 3==1`；
+
+故可以用`(2^n-1）% 3`是否等于0来等价判断思路一的条件2, 将2的幂和4的幂区分开。
 # C++
 ## 思路一
 ``` C++