Update solution 047

Author: yanglbme

README.md

@@ -1,12 +1,13 @@
+<p align="center">
+  <a href="https://github.com/yanglbme/leetcode"><img src="http://p9ucdlghd.bkt.clouddn.com/leetcode-github.png" alt="LeetCode-GitHub"></a>
+</p>
 
-![LeetCode-GitHub](http://p9ucdlghd.bkt.clouddn.com/leetcode-github.png)
+<p align="center">
+  <a href="http://makeapullrequest.com"><img src="https://img.shields.io/badge/PRs-Welcome-brightgreen.svg" alt="PRs Welcome"></a>
+  <a href="https://github.com/yanglbme/leetcode"><img src="https://img.shields.io/badge/Lang-Java%2FPython%2FJS%2FCPP%2FGo%2F...-blue.svg" alt="Language"></a>
+</p>
 
-<center>
-
-[![PRs Welcome](https://img.shields.io/badge/PRs-Welcome-brightgreen.svg)](http://makeapullrequest.com)
-[![Language](https://img.shields.io/badge/Lang-Java%2FPython%2FJS%2FCPP%2FGo%2F...-blue.svg)](https://github.com/yanglbme/leetcode)
-
-</center>
+<h2 align="center">LeetCode Solution</h2>
 
 ## Introduction
 Complete solutions to Leetcode problems, updated daily.
@@ -65,4 +66,4 @@ I'm looking for long-term contributors/partners to this repo! Send me PRs if you
 | <center> [<img src="https://avatars3.githubusercontent.com/u/21008209?v=4" width="100px;"/><br /><sub><b>yanglbme</b></sub>](https://github.com/yanglbme)<br />[💻](https://github.com/yanglbme/leetcode/commits?author=yanglbme "Code") </center> | <center> [<img src="https://avatars3.githubusercontent.com/u/23625436?v=4" width="100px;"/><br /><sub><b>chakyam</b></sub>](https://github.com/chakyam)<br />[💻](https://github.com/yanglbme/leetcode/commits?author=chakyam "Code") </center> | 
 |---|---|
 
-<!-- ALL-CONTRIBUTORS-LIST:END -->
+<!-- ALL-CONTRIBUTORS-LIST:END -->

solution/047.Permutations II/README.md

@@ -17,6 +17,8 @@
 ### 解法
 将数组的首元素依次与数组的每个元素交换(两元素不相等才进行交换)，对于每一轮交换，对后面的数组进行递归调用。当元素只剩下一个时，添加此时的数组到 list 中。
 
+注意：第 i 个数字与第 j 个数字交换时，要求[i, j) 中没有与第 j 个数字相等的数。
+
 ```java
 class Solution {
     public List<List<Integer>> permuteUnique(int[] nums) {
@@ -51,7 +53,7 @@ class Solution {
     private boolean isSwap(int[] nums, int from, int to) {
         for (int i = from; i < to; ++i) {
             if (nums[i] == nums[to]) {
-                // 相等，不进行交换
+                // [from, to) 中出现与 第 to 个数相等的数，返回 false，不进行交换和全排列操作
                 return false;
             }
         }