enable MemberName check

Author: fishercoder1534

fishercoder_checkstyle.xml

@@ -112,11 +112,11 @@
             <!--<message key="name.invalidPattern"-->
                      <!--value="Type name ''{0}'' must match pattern ''{1}''."/>-->
         <!--</module>-->
-        <!--<module name="MemberName">-->
-            <!--<property name="format" value="^[a-z][a-z0-9][a-zA-Z0-9]*$"/>-->
-            <!--<message key="name.invalidPattern"-->
-                     <!--value="Member name ''{0}'' must match pattern ''{1}''."/>-->
-        <!--</module>-->
+        <module name="MemberName">
+            <property name="format" value="^[a-z]*[a-z0-9][a-zA-Z0-9]*$"/>
+            <message key="name.invalidPattern"
+                     value="Member name ''{0}'' must match pattern ''{1}''."/>
+        </module>
         <module name="ParameterName">
             <property name="format" value="^[a-zA-Z][a-zA-Z0-9]*$"/>
             <message key="name.invalidPattern"

src/main/java/com/fishercoder/solutions/_12.java

@@ -7,7 +7,7 @@
 public class _12 {
     //looked at this post: https://discuss.leetcode.com/topic/12384/simple-solution
     public String intToRoman(int num) {
-        String M[] = new String[]{"", "M", "MM", "MMM"};
+        String M[] = new String[]{"", "m", "MM", "MMM"};
         String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
         String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
         String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};

src/main/java/com/fishercoder/solutions/_174.java

@@ -4,7 +4,7 @@
 
 /**174. Dungeon Game
 
-The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
+The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of m x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
 
 The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
 

src/main/java/com/fishercoder/solutions/_466.java

@@ -8,7 +8,7 @@
  On the other hand, we define that string s1 can be obtained from string s2 if we can remove some characters from s2 such that it becomes s1.
  For example, “abc” can be obtained from “abdbec” based on our definition, but it can not be obtained from “acbbe”.
  You are given two non-empty strings s1 and s2 (each at most 100 characters long) and two integers 0 ≤ n1 ≤ 106 and 1 ≤ n2 ≤ 106.
- Now consider the strings S1 and S2, where S1=[s1,n1] and S2=[s2,n2]. Find the maximum integer M such that [S2,M] can be obtained from S1.
+ Now consider the strings S1 and S2, where S1=[s1,n1] and S2=[s2,n2]. Find the maximum integer m such that [S2,m] can be obtained from S1.
 
  Example:
 

src/main/java/com/fishercoder/solutions/_482.java

@@ -1,7 +1,7 @@
 package com.fishercoder.solutions;
 
 /**
- * Now you are given a string S, which represents a software license key which we would like to format. The string S is composed of alphanumerical characters and dashes. The dashes split the alphanumerical characters within the string into groups. (i.e. if there are M dashes, the string is split into M+1 groups). The dashes in the given string are possibly misplaced.
+ * Now you are given a string S, which represents a software license key which we would like to format. The string S is composed of alphanumerical characters and dashes. The dashes split the alphanumerical characters within the string into groups. (i.e. if there are m dashes, the string is split into m+1 groups). The dashes in the given string are possibly misplaced.
 
  We want each group of characters to be of length K (except for possibly the first group, which could be shorter, but still must contain at least one character). To satisfy this requirement, we will reinsert dashes. Additionally, all the lower case letters in the string must be converted to upper case.
 

src/main/java/com/fishercoder/solutions/_488.java

@@ -41,20 +41,20 @@ public class _488 {
     /**credit: https://discuss.leetcode.com/topic/79820/short-java-solution-beats-98
      * Two layer of recursion, pretty cool!*/
 
-    int MAXCOUNT = 6;   // the max balls you need will not exceed 6 since "The number of balls in your hand won't exceed 5"
+    int maxcount = 6;   // the max balls you need will not exceed 6 since "The number of balls in your hand won't exceed 5"
 
     public int findMinStep(String board, String hand) {
         int[] handCount = new int[26];
         for (int i = 0; i < hand.length(); ++i) {
             ++handCount[hand.charAt(i) - 'A'];
         }
         int result = dfs(board + "#", handCount);  // append a "#" to avoid special process while j==board.length, make the code shorter.
-        return result == MAXCOUNT ? -1 : result;
+        return result == maxcount ? -1 : result;
     }
     private int dfs(String s, int[] handCount) {
         s = removeConsecutive(s);
         if (s.equals("#")) return 0;
-        int result = MAXCOUNT;
+        int result = maxcount;
         int need = 0;
         for (int i = 0, j = 0 ; j < s.length(); ++j) {
             if (s.charAt(j) == s.charAt(i)) continue;

src/main/java/com/fishercoder/solutions/_498.java

@@ -3,7 +3,7 @@
 /**
  * 498. Diagonal Traverse
  *
- *  Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order
+ *  Given a matrix of m x N elements (m rows, N columns), return all elements of the matrix in diagonal order
  *  as shown in the below image.
 
  Example:

src/main/java/com/fishercoder/solutions/_529.java

@@ -6,19 +6,19 @@
 /**
  * Let's play the minesweeper game (Wikipedia, online game)!
 
- You are given a 2D char matrix representing the game board. 'M' represents an unrevealed mine, 'E' represents an unrevealed empty square, 'B' represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit ('1' to '8') represents how many mines are adjacent to this revealed square, and finally 'X' represents a revealed mine.
+ You are given a 2D char matrix representing the game board. 'm' represents an unrevealed mine, 'E' represents an unrevealed empty square, 'B' represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit ('1' to '8') represents how many mines are adjacent to this revealed square, and finally 'X' represents a revealed mine.
 
- Now given the next click position (row and column indices) among all the unrevealed squares ('M' or 'E'), return the board after revealing this position according to the following rules:
+ Now given the next click position (row and column indices) among all the unrevealed squares ('m' or 'E'), return the board after revealing this position according to the following rules:
 
- If a mine ('M') is revealed, then the game is over - change it to 'X'.
+ If a mine ('m') is revealed, then the game is over - change it to 'X'.
  If an empty square ('E') with no adjacent mines is revealed, then change it to revealed blank ('B') and all of its adjacent unrevealed squares should be revealed recursively.
  If an empty square ('E') with at least one adjacent mine is revealed, then change it to a digit ('1' to '8') representing the number of adjacent mines.
  Return the board when no more squares will be revealed.
  Example 1:
  Input:
 
  [['E', 'E', 'E', 'E', 'E'],
- ['E', 'E', 'M', 'E', 'E'],
+ ['E', 'E', 'm', 'E', 'E'],
  ['E', 'E', 'E', 'E', 'E'],
  ['E', 'E', 'E', 'E', 'E']]
 
@@ -27,7 +27,7 @@ If an empty square ('E') with at least one adjacent mine is revealed, then chang
  Output:
 
  [['B', '1', 'E', '1', 'B'],
- ['B', '1', 'M', '1', 'B'],
+ ['B', '1', 'm', '1', 'B'],
  ['B', '1', '1', '1', 'B'],
  ['B', 'B', 'B', 'B', 'B']]
 
@@ -37,7 +37,7 @@ If an empty square ('E') with at least one adjacent mine is revealed, then chang
  Input:
 
  [['B', '1', 'E', '1', 'B'],
- ['B', '1', 'M', '1', 'B'],
+ ['B', '1', 'm', '1', 'B'],
  ['B', '1', '1', '1', 'B'],
  ['B', 'B', 'B', 'B', 'B']]
 
@@ -54,7 +54,7 @@ If an empty square ('E') with at least one adjacent mine is revealed, then chang
 
  Note:
  The range of the input matrix's height and width is [1,50].
- The click position will only be an unrevealed square ('M' or 'E'), which also means the input board contains at least one clickable square.
+ The click position will only be an unrevealed square ('m' or 'E'), which also means the input board contains at least one clickable square.
  The input board won't be a stage when game is over (some mines have been revealed).
  For simplicity, not mentioned rules should be ignored in this problem. For example, you don't need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.
  */

src/main/java/com/fishercoder/solutions/_547.java

@@ -8,8 +8,8 @@
  * For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C.
  * And we defined a friend circle is a group of students who are direct or indirect friends.
 
- Given a N*N matrix M representing the friend relationship between students in the class.
- If M[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not.
+ Given a N*N matrix m representing the friend relationship between students in the class.
+ If m[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not.
  And you have to output the total number of friend circles among all the students.
 
  Example 1:
@@ -32,8 +32,8 @@
 
  Note:
  N is in range [1,200].
- M[i][i] = 1 for all students.
- If M[i][j] = 1, then M[j][i] = 1.
+ m[i][i] = 1 for all students.
+ If m[i][j] = 1, then m[j][i] = 1.
  */
 public class _547 {
 

src/main/java/com/fishercoder/solutions/_562.java

@@ -6,7 +6,7 @@
 
 /**Longest Line of Consecutive One in Matrix
  *
- * Given a 01 matrix M, find the longest line of consecutive one in the matrix. The line could be horizontal, vertical, diagonal or anti-diagonal.
+ * Given a 01 matrix m, find the longest line of consecutive one in the matrix. The line could be horizontal, vertical, diagonal or anti-diagonal.
 
  Example:
 

src/main/java/com/fishercoder/solutions/_598.java

@@ -3,10 +3,10 @@
 /**
  * 598. Range Addition II
 
- Given an m * n matrix M initialized with all 0's and several update operations.
+ Given an m * n matrix m initialized with all 0's and several update operations.
  Operations are represented by a 2D array,
  and each operation is represented by an array with two positive integers a and b,
- which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.
+ which means m[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.
 
  You need to count and return the number of maximum integers in the matrix after performing all the operations.
 
@@ -17,22 +17,22 @@
  Output: 4
 
  Explanation:
- Initially, M =
+ Initially, m =
  [[0, 0, 0],
  [0, 0, 0],
  [0, 0, 0]]
 
- After performing [2,2], M =
+ After performing [2,2], m =
  [[1, 1, 0],
  [1, 1, 0],
  [0, 0, 0]]
 
- After performing [3,3], M =
+ After performing [3,3], m =
  [[2, 2, 1],
  [2, 2, 1],
  [1, 1, 1]]
 
- So the maximum integer in M is 2, and there are four of it in M. So return 4.
+ So the maximum integer in m is 2, and there are four of it in m. So return 4.
  Note:
  The range of m and n is [1,40000].
  The range of a is [1,m], and the range of b is [1,n].

src/main/java/com/fishercoder/solutions/_631.java

@@ -82,7 +82,7 @@ public class _631 {
     /**Credit: https://leetcode.com/articles/design-excel-sum-formula/#approach-1-using-topological-sortaccepted*/
     public static class Excel {
 
-        Formula[][] Formulas;
+        Formula[][] formulas;
 
         class Formula {
             Formula(HashMap<String, Integer> c, int v) {
@@ -97,18 +97,18 @@ class Formula {
         Stack<int[]> stack = new Stack<>();
 
         public Excel(int H, char W) {
-            Formulas = new Formula[H][(W - 'A') + 1];
+            formulas = new Formula[H][(W - 'A') + 1];
         }
 
         public int get(int r, char c) {
-            if (Formulas[r - 1][c - 'A'] == null) {
+            if (formulas[r - 1][c - 'A'] == null) {
                 return 0;
             }
-            return Formulas[r - 1][c - 'A'].val;
+            return formulas[r - 1][c - 'A'].val;
         }
 
         public void set(int r, char c, int v) {
-            Formulas[r - 1][c - 'A'] = new Formula(new HashMap<String, Integer>(), v);
+            formulas[r - 1][c - 'A'] = new Formula(new HashMap<String, Integer>(), v);
             topologicalSort(r - 1, c - 'A');
             execute_stack();
         }
@@ -117,14 +117,14 @@ public int sum(int r, char c, String[] strs) {
             HashMap<String, Integer> cells = convert(strs);
             int summ = calculate_sum(r - 1, c - 'A', cells);
             set(r, c, summ);
-            Formulas[r - 1][c - 'A'] = new Formula(cells, summ);
+            formulas[r - 1][c - 'A'] = new Formula(cells, summ);
             return summ;
         }
 
         public void topologicalSort(int r, int c) {
-            for (int i = 0; i < Formulas.length; i++) {
-                for (int j = 0; j < Formulas[0].length; j++) {
-                    if (Formulas[i][j] != null && Formulas[i][j].cells.containsKey("" + (char) ('A' + c) + (r + 1))) {
+            for (int i = 0; i < formulas.length; i++) {
+                for (int j = 0; j < formulas[0].length; j++) {
+                    if (formulas[i][j] != null && formulas[i][j].cells.containsKey("" + (char) ('A' + c) + (r + 1))) {
                         topologicalSort(i, j);
                     }
                 }
@@ -135,8 +135,8 @@ public void topologicalSort(int r, int c) {
         public void execute_stack() {
             while (!stack.isEmpty()) {
                 int[] top = stack.pop();
-                if (Formulas[top[0]][top[1]].cells.size() > 0) {
-                    calculate_sum(top[0], top[1], Formulas[top[0]][top[1]].cells);
+                if (formulas[top[0]][top[1]].cells.size() > 0) {
+                    calculate_sum(top[0], top[1], formulas[top[0]][top[1]].cells);
                 }
             }
         }
@@ -164,9 +164,9 @@ public int calculate_sum(int r, int c, HashMap<String, Integer> cells) {
             int sum = 0;
             for (String s : cells.keySet()) {
                 int x = Integer.parseInt(s.substring(1)) - 1, y = s.charAt(0) - 'A';
-                sum += (Formulas[x][y] != null ? Formulas[x][y].val : 0) * cells.get(s);
+                sum += (formulas[x][y] != null ? formulas[x][y].val : 0) * cells.get(s);
             }
-            Formulas[r][c] = new Formula(cells, sum);
+            formulas[r][c] = new Formula(cells, sum);
             return sum;
         }
     }

src/main/java/com/fishercoder/solutions/_639.java

@@ -32,7 +32,7 @@
  */
 public class _639 {
     /**reference: https://leetcode.com/articles/decode-ways-ii/#approach-2-dynamic-programming-accepted*/
-    int M = 1000000007;
+    int m = 1000000007;
     public int numDecodings(String s) {
         long[] dp = new long[s.length() + 1];
         dp[0] = 1;
@@ -41,20 +41,20 @@ public int numDecodings(String s) {
             if (s.charAt(i) == '*') {
                 dp[i + 1] = 9 * dp[i];
                 if (s.charAt(i - 1) == '1') {
-                    dp[i + 1] = (dp[i + 1] + 9 * dp[i - 1]) % M;
+                    dp[i + 1] = (dp[i + 1] + 9 * dp[i - 1]) % m;
                 } else if (s.charAt(i - 1) == '2') {
-                    dp[i + 1] = (dp[i + 1] + 6 * dp[i - 1]) % M;
+                    dp[i + 1] = (dp[i + 1] + 6 * dp[i - 1]) % m;
                 } else if (s.charAt(i - 1) == '*') {
-                    dp[i + 1] = (dp[i + 1] + 15 * dp[i - 1]) % M;
+                    dp[i + 1] = (dp[i + 1] + 15 * dp[i - 1]) % m;
                 }
             } else {
                 dp[i + 1] = s.charAt(i) != '0' ? dp[i] : 0;
                 if (s.charAt(i - 1) == '1') {
-                    dp[i + 1] = (dp[i + 1] + dp[i - 1]) % M;
+                    dp[i + 1] = (dp[i + 1] + dp[i - 1]) % m;
                 } else if (s.charAt(i - 1) == '2' && s.charAt(i) <= '6') {
-                    dp[i + 1] = (dp[i + 1] + dp[i - 1]) % M;
+                    dp[i + 1] = (dp[i + 1] + dp[i - 1]) % m;
                 } else if (s.charAt(i - 1) == '*') {
-                    dp[i + 1] = (dp[i + 1] + (s.charAt(i) <= '6' ? 2 : 1) * dp[i - 1]) % M;
+                    dp[i + 1] = (dp[i + 1] + (s.charAt(i) <= '6' ? 2 : 1) * dp[i - 1]) % m;
                 }
             }
         }

src/main/java/com/fishercoder/solutions/_661.java

@@ -3,7 +3,7 @@
 /**
  * 661. Image Smoother
  *
- * Given a 2D integer matrix M representing the gray scale of an image,
+ * Given a 2D integer matrix m representing the gray scale of an image,
  * you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of
  * all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.