diff --git a/notes/10.1 斐波那契数列.md b/notes/10.1 斐波那契数列.md
index 63fd6d7f..fd9e836b 100644
--- a/notes/10.1 斐波那契数列.md	
+++ b/notes/10.1 斐波那契数列.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[NowCoder](https://www.nowcoder.com/practice/c6c7742f5ba7442aada113136ddea0c3?tpId=13&tqId=11160&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/c6c7742f5ba7442aada113136ddea0c3?tpId=13&tqId=11160&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
 
 ## 题目描述
 
diff --git a/notes/10.2 矩形覆盖.md b/notes/10.2 矩形覆盖.md
index 56e244a5..01e833f7 100644
--- a/notes/10.2 矩形覆盖.md	
+++ b/notes/10.2 矩形覆盖.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[NowCoder](https://www.nowcoder.com/practice/72a5a919508a4251859fb2cfb987a0e6?tpId=13&tqId=11163&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/72a5a919508a4251859fb2cfb987a0e6?tpId=13&tqId=11163&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
 
 ## 题目描述
 
@@ -27,7 +27,7 @@
 <div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/508c6e52-9f93-44ed-b6b9-e69050e14807.jpg" width="370px"> </div><br>
 
 ```java
-public int RectCover(int n) {
+public int rectCover(int n) {
     if (n <= 2)
         return n;
     int pre2 = 1, pre1 = 2;
diff --git a/notes/10.3 跳台阶.md b/notes/10.3 跳台阶.md
index 4cf0ee43..2072cb90 100644
--- a/notes/10.3 跳台阶.md	
+++ b/notes/10.3 跳台阶.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[NowCoder](https://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4?tpId=13&tqId=11161&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4?tpId=13&tqId=11161&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
 
 ## 题目描述
 
diff --git a/notes/10.4 变态跳台阶.md b/notes/10.4 变态跳台阶.md
index 605ce196..f6099a17 100644
--- a/notes/10.4 变态跳台阶.md	
+++ b/notes/10.4 变态跳台阶.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[NowCoder](https://www.nowcoder.com/practice/22243d016f6b47f2a6928b4313c85387?tpId=13&tqId=11162&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/22243d016f6b47f2a6928b4313c85387?tpId=13&tqId=11162&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
 
 ## 题目描述
 
@@ -15,7 +15,7 @@
 ### 动态规划
 
 ```java
-public int JumpFloorII(int target) {
+public int jumpFloorII(int target) {
     int[] dp = new int[target];
     Arrays.fill(dp, 1);
     for (int i = 1; i < target; i++)
diff --git a/notes/12. 矩阵中的路径.md b/notes/12. 矩阵中的路径.md
index 913d9fe3..15778537 100644
--- a/notes/12. 矩阵中的路径.md	
+++ b/notes/12. 矩阵中的路径.md	
@@ -1,6 +1,6 @@
 # 12. 矩阵中的路径
 
-[NowCoder](https://www.nowcoder.com/practice/c61c6999eecb4b8f88a98f66b273a3cc?tpId=13&tqId=11218&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/69fe7a584f0a445da1b6652978de5c38?tpId=13&tqId=11218&tab=answerKey&from=cyc_github)
 
 ## 题目描述
 
@@ -19,46 +19,60 @@
 本题的输入是数组而不是矩阵（二维数组），因此需要先将数组转换成矩阵。
 
 ```java
-private final static int[][] next = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
-private int rows;
-private int cols;
+public class Solution {
+    private final static int[][] next = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
+    private int rows;
+    private int cols;
 
-public boolean hasPath(char[] array, int rows, int cols, char[] str) {
-    if (rows == 0 || cols == 0) return false;
-    this.rows = rows;
-    this.cols = cols;
-    boolean[][] marked = new boolean[rows][cols];
-    char[][] matrix = buildMatrix(array);
-    for (int i = 0; i < rows; i++)
-        for (int j = 0; j < cols; j++)
-            if (backtracking(matrix, str, marked, 0, i, j))
-                return true;
-
-    return false;
-}
-
-private boolean backtracking(char[][] matrix, char[] str,
-                             boolean[][] marked, int pathLen, int r, int c) {
-
-    if (pathLen == str.length) return true;
-    if (r < 0 || r >= rows || c < 0 || c >= cols
-            || matrix[r][c] != str[pathLen] || marked[r][c]) {
+    public boolean hasPath (String val, int rows, int cols, String path) {
+        if (rows == 0 || cols == 0) return false;
+        this.rows = rows;
+        this.cols = cols;
+        char[] array = val.toCharArray();
+        char[][] matrix = buildMatrix(array);
+        char[] pathList = path.toCharArray();
+        boolean[][] marked = new boolean[rows][cols];
+        for (int i = 0; i < rows; i++)
+            for (int j = 0; j < cols; j++)
+                if (backtracking(matrix, pathList, marked, 0, i, j))
+                    return true;
 
         return false;
     }
-    marked[r][c] = true;
-    for (int[] n : next)
-        if (backtracking(matrix, str, marked, pathLen + 1, r + n[0], c + n[1]))
-            return true;
-    marked[r][c] = false;
-    return false;
-}
 
-private char[][] buildMatrix(char[] array) {
-    char[][] matrix = new char[rows][cols];
-    for (int r = 0, idx = 0; r < rows; r++)
-        for (int c = 0; c < cols; c++)
-            matrix[r][c] = array[idx++];
-    return matrix;
+    private boolean backtracking(char[][] matrix, char[] pathList,
+                                 boolean[][] marked, int pathLen, int r, int c) {
+
+        if (pathLen == pathList.length) return true;
+        if (r < 0 || r >= rows || c < 0 || c >= cols
+                || matrix[r][c] != pathList[pathLen] || marked[r][c]) {
+
+            return false;
+        }
+        marked[r][c] = true;
+        for (int[] n : next)
+            if (backtracking(matrix, pathList, marked, pathLen + 1, r + n[0], c + n[1]))
+                return true;
+        marked[r][c] = false;
+        return false;
+    }
+
+    private char[][] buildMatrix(char[] array) {
+        char[][] matrix = new char[rows][cols];
+        for (int r = 0, idx = 0; r < rows; r++)
+            for (int c = 0; c < cols; c++)
+                matrix[r][c] = array[idx++];
+        return matrix;
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        String val = "ABCESFCSADEE";
+        int rows = 3;
+        int cols = 4;
+        String path = "ABCCED";
+        boolean res = solution.hasPath(val, rows, cols, path);
+        System.out.println(res);
+    }
 }
 ```
diff --git a/notes/13. 机器人的运动范围.md b/notes/13. 机器人的运动范围.md
index 766b1bd6..dd96086e 100644
--- a/notes/13. 机器人的运动范围.md	
+++ b/notes/13. 机器人的运动范围.md	
@@ -1,6 +1,6 @@
 # 13. 机器人的运动范围
 
-[NowCoder](https://www.nowcoder.com/practice/6e5207314b5241fb83f2329e89fdecc8?tpId=13&tqId=11219&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/6e5207314b5241fb83f2329e89fdecc8?tpId=13&tqId=11219&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
 
 ## 题目描述
 
diff --git a/notes/14. 剪绳子.md b/notes/14. 剪绳子.md
index 16ca2493..9108b635 100644
--- a/notes/14. 剪绳子.md	
+++ b/notes/14. 剪绳子.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[Leetcode](https://leetcode.com/problems/integer-break/description/)
+[牛客网](https://www.nowcoder.com/practice/57d85990ba5b440ab888fc72b0751bf8?tpId=13&tqId=33257&tab=answerKey&from=cyc_github)
 
 ## 题目描述
 
@@ -37,7 +37,7 @@ return 36 (10 = 3 + 3 + 4)
 继续拆成更大的绳子可以发现都比拆成 2 和 3 的效果更差，因此我们只考虑将绳子拆成 2 和 3，并且优先拆成 3，当拆到绳子长度 n 等于 4 时，也就是出现 3+1，此时只能拆成 2+2。
 
 ```java
-public int integerBreak(int n) {
+public int cutRope(int n) {
     if (n < 2)
         return 0;
     if (n == 2)
@@ -55,7 +55,7 @@ public int integerBreak(int n) {
 ### 动态规划
 
 ```java
-public int integerBreak(int n) {
+public int cutRope(int n) {
     int[] dp = new int[n + 1];
     dp[1] = 1;
     for (int i = 2; i <= n; i++)
diff --git a/notes/18.2 删除链表中重复的结点.md b/notes/18.2 删除链表中重复的结点.md
index 57e247bf..5d1f871b 100644
--- a/notes/18.2 删除链表中重复的结点.md	
+++ b/notes/18.2 删除链表中重复的结点.md	
@@ -1,6 +1,6 @@
 # 18.2 删除链表中重复的结点
 
-[NowCoder](https://www.nowcoder.com/practice/fc533c45b73a41b0b44ccba763f866ef?tpId=13&tqId=11209&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/fc533c45b73a41b0b44ccba763f866ef?tpId=13&tqId=11209&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
 
 ## 题目描述
 
diff --git a/notes/19. 正则表达式匹配.md b/notes/19. 正则表达式匹配.md
index 97de47d7..a0ad0053 100644
--- a/notes/19. 正则表达式匹配.md	
+++ b/notes/19. 正则表达式匹配.md	
@@ -1,6 +1,6 @@
 # 19. 正则表达式匹配
 
-[NowCoder](https://www.nowcoder.com/practice/45327ae22b7b413ea21df13ee7d6429c?tpId=13&tqId=11205&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/28970c15befb4ff3a264189087b99ad4?tpId=13&tqId=11205&tab=answerKey&from=cyc_github)
 
 ## 题目描述
 
@@ -13,22 +13,22 @@
 应该注意到，'.' 是用来当做一个任意字符，而 '\*' 是用来重复前面的字符。这两个的作用不同，不能把 '.' 的作用和 '\*' 进行类比，从而把它当成重复前面字符一次。
 
 ```java
-public boolean match(char[] str, char[] pattern) {
+public boolean match(String str, String pattern) {
 
-    int m = str.length, n = pattern.length;
+    int m = str.length(), n = pattern.length();
     boolean[][] dp = new boolean[m + 1][n + 1];
 
     dp[0][0] = true;
     for (int i = 1; i <= n; i++)
-        if (pattern[i - 1] == '*')
+        if (pattern.charAt(i - 1) == '*')
             dp[0][i] = dp[0][i - 2];
 
     for (int i = 1; i <= m; i++)
         for (int j = 1; j <= n; j++)
-            if (str[i - 1] == pattern[j - 1] || pattern[j - 1] == '.')
+            if (str.charAt(i - 1) == pattern.charAt(j - 1) || pattern.charAt(j - 1) == '.')
                 dp[i][j] = dp[i - 1][j - 1];
-            else if (pattern[j - 1] == '*')
-                if (pattern[j - 2] == str[i - 1] || pattern[j - 2] == '.') {
+            else if (pattern.charAt(j - 1) == '*')
+                if (pattern.charAt(j - 2) == str.charAt(i - 1) || pattern.charAt(j - 2) == '.') {
                     dp[i][j] |= dp[i][j - 1]; // a* counts as single a
                     dp[i][j] |= dp[i - 1][j]; // a* counts as multiple a
                     dp[i][j] |= dp[i][j - 2]; // a* counts as empty
diff --git a/notes/20. 表示数值的字符串.md b/notes/20. 表示数值的字符串.md
index c03369ba..da4b4bf6 100644
--- a/notes/20. 表示数值的字符串.md	
+++ b/notes/20. 表示数值的字符串.md	
@@ -1,6 +1,6 @@
 # 20. 表示数值的字符串
 
-[NowCoder](https://www.nowcoder.com/practice/6f8c901d091949a5837e24bb82a731f2?tpId=13&tqId=11206&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/e69148f8528c4039ad89bb2546fd4ff8?tpId=13&tqId=11206&tab=answerKey&from=cyc_github)
 
 ## 题目描述
 
@@ -41,8 +41,8 @@ false
 ```
 
 ```java
-public boolean isNumeric(char[] str) {
-    if (str == null || str.length == 0)
+public boolean isNumeric (String str) {
+    if (str == null || str.length() == 0)
         return false;
     return new String(str).matches("[+-]?\\d*(\\.\\d+)?([eE][+-]?\\d+)?");
 }
diff --git a/notes/21. 调整数组顺序使奇数位于偶数前面.md b/notes/21. 调整数组顺序使奇数位于偶数前面.md
index 7b2e1699..13ca98e5 100644
--- a/notes/21. 调整数组顺序使奇数位于偶数前面.md	
+++ b/notes/21. 调整数组顺序使奇数位于偶数前面.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/beb5aa231adc45b2a5dcc5b62c93f593?tpId=13&tqId=11166&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/ef1f53ef31ca408cada5093c8780f44b?tpId=13&tqId=11166&tab=answerKey&from=cyc_github)
 
 ## 题目描述
 
@@ -15,7 +15,7 @@
 方法一：创建一个新数组，时间复杂度 O(N)，空间复杂度 O(N)。
 
 ```java
-public void reOrderArray(int[] nums) {
+public int[] reOrderArray (int[] nums) {
     // 奇数个数
     int oddCnt = 0;
     for (int x : nums)
@@ -29,6 +29,7 @@ public void reOrderArray(int[] nums) {
         else
             nums[j++] = num;
     }
+    return nums;
 }
 
 private boolean isEven(int x) {
@@ -39,7 +40,7 @@ private boolean isEven(int x) {
 方法二：使用冒泡思想，每次都将当前偶数上浮到当前最右边。时间复杂度 O(N<sup>2</sup>)，空间复杂度 O(1)，时间换空间。
 
 ```java
-public void reOrderArray(int[] nums) {
+public int[] reOrderArray(int[] nums) {
     int N = nums.length;
     for (int i = N - 1; i > 0; i--) {
         for (int j = 0; j < i; j++) {
@@ -48,6 +49,7 @@ public void reOrderArray(int[] nums) {
             }
         }
     }
+    return nums;
 }
 
 private boolean isEven(int x) {
diff --git a/notes/22. 链表中倒数第 K 个结点.md b/notes/22. 链表中倒数第 K 个结点.md
index f77964e7..aa00d274 100644
--- a/notes/22. 链表中倒数第 K 个结点.md	
+++ b/notes/22. 链表中倒数第 K 个结点.md	
@@ -1,6 +1,6 @@
 # 22. 链表中倒数第 K 个结点
 
-[NowCoder](https://www.nowcoder.com/practice/529d3ae5a407492994ad2a246518148a?tpId=13&tqId=11167&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/886370fe658f41b498d40fb34ae76ff9?tpId=13&tqId=11167&tab=answerKey&from=cyc_github)
 
 ## 解题思路
 
diff --git a/notes/27. 二叉树的镜像.md b/notes/27. 二叉树的镜像.md
index f9ca746a..18b5f0fa 100644
--- a/notes/27. 二叉树的镜像.md	
+++ b/notes/27. 二叉树的镜像.md	
@@ -1,6 +1,6 @@
 # 27. 二叉树的镜像
 
-[NowCoder](https://www.nowcoder.com/practice/564f4c26aa584921bc75623e48ca3011?tpId=13&tqId=11171&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/a9d0ecbacef9410ca97463e4a5c83be7?tpId=13&tqId=11171&tab=answerKey&from=cyc_github)
 
 ## 题目描述
 
@@ -9,12 +9,13 @@
 ## 解题思路
 
 ```java
-public void Mirror(TreeNode root) {
+public TreeNode Mirror(TreeNode root) {
     if (root == null)
-        return;
+        return root;
     swap(root);
     Mirror(root.left);
     Mirror(root.right);
+    return root;
 }
 
 private void swap(TreeNode root) {
diff --git a/notes/5. 替换空格.md b/notes/5. 替换空格.md
index 211ae0c6..bd808616 100644
--- a/notes/5. 替换空格.md	
+++ b/notes/5. 替换空格.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/4060ac7e3e404ad1a894ef3e17650423?tpId=13&tqId=11155&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/0e26e5551f2b489b9f58bc83aa4b6c68?tpId=13&tqId=11155&tab=answerKey&from=cyc_github)
 
 ## 题目描述
 
diff --git a/notes/56. 数组中只出现一次的数字.md b/notes/56. 数组中只出现一次的数字.md
index 21c1c3dc..4ab74ad3 100644
--- a/notes/56. 数组中只出现一次的数字.md	
+++ b/notes/56. 数组中只出现一次的数字.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/e02fdb54d7524710a7d664d082bb7811?tpId=13&tqId=11193&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[牛客网](https://www.nowcoder.com/practice/389fc1c3d3be4479a154f63f495abff8?tpId=13&tqId=11193&tab=answerKey&from=cyc_github)
 
 ## 题目描述
 
@@ -19,16 +19,27 @@
 下面的解法中，num1 和 num2 数组的第一个元素是用来保持返回值的... 实际开发中不推荐这种返回值的方式。
 
 ```java
-public void FindNumsAppearOnce(int[] nums, int num1[], int num2[]) {
+public int[] FindNumsAppearOnce (int[] nums) {
+    int[] res = new int[2];
     int diff = 0;
     for (int num : nums)
         diff ^= num;
     diff &= -diff;
     for (int num : nums) {
         if ((num & diff) == 0)
-            num1[0] ^= num;
+            res[0] ^= num;
         else
-            num2[0] ^= num;
+            res[1] ^= num;
     }
+    if (res[0] > res[1]) {
+        swap(res);
+    }
+    return res;
+}
+
+private void swap(int[] nums) {
+    int t = nums[0];
+    nums[0] = nums[1];
+    nums[1] = t;
 }
 ```