refactor for format

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_162.java

@@ -20,41 +20,43 @@ public class _162 {
      * or nums[right] > nums[right+1], then we could return right as the result
      */
     public static int findPeakElement_Ologn(int[] nums) {
-        
-        if(nums == null || nums.length == 0) return 0;
-        int left = 0, right = nums.length-1;
-        while(left+1 < right){
-            int mid = left + (right-left)/2;
-            if(nums[mid] < nums[mid+1]){
+
+        if (nums == null || nums.length == 0) return 0;
+        int left = 0, right = nums.length - 1;
+        while (left + 1 < right) {
+            int mid = left + (right - left) / 2;
+            if (nums[mid] < nums[mid + 1]) {
                 left = mid;
             } else {
                 right = mid;
             }
         }
-        return (left == nums.length-1 || nums[left] > nums[left+1]) ? left : right;
-    
+        return (left == nums.length - 1 || nums[left] > nums[left + 1]) ? left : right;
+
     }
 
-    public static void main(String...strings){
+    public static void main(String... strings) {
 //        int[] nums = new int[]{1,2};
 //        int[] nums = new int[]{1};
-        int[] nums = new int[]{1,2,3,1};
+        int[] nums = new int[]{1, 2, 3, 1};
 //        System.out.println(findPeakElement(nums));
         System.out.println(findPeakElement_Ologn(nums));
     }
-    
-    /**My original O(n) solution.*/
+
+    /**
+     * My original O(n) solution.
+     */
     public static int findPeakElement(int[] nums) {
-        if(nums == null || nums.length == 0) return 0;
+        if (nums == null || nums.length == 0) return 0;
         int n = nums.length, result = 0;
-        for(int i = 0; i < n; i++){
-            if(i == 0 && n > 1 && nums[i] > nums[i+1]){
+        for (int i = 0; i < n; i++) {
+            if (i == 0 && n > 1 && nums[i] > nums[i + 1]) {
                 result = i;
                 break;
-            } else if(i == n-1 && i > 0 && nums[i] > nums[i-1]){
+            } else if (i == n - 1 && i > 0 && nums[i] > nums[i - 1]) {
                 result = i;
                 break;
-            } else if(i > 0 && i < n-1 && nums[i] > nums[i-1] && nums[i] > nums[i+1]){
+            } else if (i > 0 && i < n - 1 && nums[i] > nums[i - 1] && nums[i] > nums[i + 1]) {
                 result = i;
                 break;
             }

src/main/java/com/fishercoder/solutions/_164.java

@@ -32,7 +32,6 @@ public int maximumGap(int[] nums) {
     }
 
 
-
     //http://www.programcreek.com/2014/03/leetcode-maximum-gap-java/
     class Bucket {
         int min = -1;
@@ -61,7 +60,7 @@ public int maximumGap_from_programcreek_1(int[] nums) {
             buckets[i] = new Bucket();
         }
 
-        double interval = (double) nums.length/(maxNum - minNum);
+        double interval = (double) nums.length / (maxNum - minNum);
         //distribute the array to different buckets
         for (int i = 0; i < nums.length; i++) {
             int index = (int) ((nums[i] - minNum) * interval);
@@ -104,10 +103,10 @@ public int maximumGap_from_programcreek_2(int[] nums) {
             buckets[i] = new Bucket();
         }
 
-        double gap = (double) (maxNum - minNum)/(nums.length-1);
+        double gap = (double) (maxNum - minNum) / (nums.length - 1);
         //distribute the array to different buckets
         for (int i = 0; i < nums.length; i++) {
-            int index = (int) ((nums[i] - minNum)/gap);
+            int index = (int) ((nums[i] - minNum) / gap);
             if (buckets[index].min == -1) {
                 buckets[index].min = nums[i];
                 buckets[index].max = nums[i];

src/main/java/com/fishercoder/solutions/_165.java

@@ -16,35 +16,35 @@ public static int compareVersion(String version1, String version2) {
         String[] v1s = version1.split("\\.");//escaping it is very important! Otherwise, it's not going to work as expected!
         String[] v2s = version2.split("\\.");
         int len = (v1s.length < v2s.length) ? v2s.length : v1s.length;
-        for(int i = 0; i < len; i++){
-            if(v1s.length == i){
-                while(i < len){
-                    if(Integer.parseInt(v2s[i]) > 0) return -1;
+        for (int i = 0; i < len; i++) {
+            if (v1s.length == i) {
+                while (i < len) {
+                    if (Integer.parseInt(v2s[i]) > 0) return -1;
                     i++;
                 }
-            } else if(v2s.length == i){
-                while(i < len){
-                    if(Integer.parseInt(v1s[i]) > 0) return 1;
+            } else if (v2s.length == i) {
+                while (i < len) {
+                    if (Integer.parseInt(v1s[i]) > 0) return 1;
                     i++;
                 }
             } else {
-                if(Integer.parseInt(v1s[i]) > Integer.parseInt(v2s[i])) return 1;
-                else if(Integer.parseInt(v2s[i]) > Integer.parseInt(v1s[i])) return -1;
+                if (Integer.parseInt(v1s[i]) > Integer.parseInt(v2s[i])) return 1;
+                else if (Integer.parseInt(v2s[i]) > Integer.parseInt(v1s[i])) return -1;
             }
         }
         return 0;
     }
-    
-    public static void main(String...args){
+
+    public static void main(String... args) {
 //        String version1 = "1.1";
 //        String version2 = "1.2";//should return -1
-        
+
 //        String version1 = "1.0.1";
 //        String version2 = "1";//should return 1
-        
+
         String version1 = "1.0";
         String version2 = "1";//should return 0
-        
+
         /**"1.0.1", "1"*/
         System.out.println(compareVersion(version1, version2));
     }

src/main/java/com/fishercoder/solutions/_166.java

@@ -22,24 +22,24 @@ public class _166 {
     public String fractionToDecimal(int numerator, int denominator) {
         String sign = (numerator >= 0 && denominator >= 0) || (numerator < 0 && denominator < 0) ? "" : "-";
         if (numerator == 0) return "0";
-        long num = Math.abs((long)numerator);
-        long deno = Math.abs((long)denominator);
+        long num = Math.abs((long) numerator);
+        long deno = Math.abs((long) denominator);
         StringBuilder stringBuilder = new StringBuilder();
         stringBuilder.append(sign);
-        long integral = Math.abs(num/deno);
+        long integral = Math.abs(num / deno);
         stringBuilder.append(integral);
-        if (numerator%denominator == 0) {
+        if (numerator % denominator == 0) {
             return stringBuilder.toString();
         } else {
             stringBuilder.append(".");
         }
-        long remainder = num%deno;
+        long remainder = num % deno;
 
         Map<Long, Integer> map = new HashMap<>();
         while (!map.containsKey(remainder)) {
             map.put(remainder, stringBuilder.length());
-            long n = remainder*10/deno;
-            remainder = remainder*10%deno;
+            long n = remainder * 10 / deno;
+            remainder = remainder * 10 % deno;
             if (remainder != 0 || (remainder == 0 && !map.containsKey(remainder))) {
                 stringBuilder.append(n);
             }

src/main/java/com/fishercoder/solutions/_168.java

@@ -17,16 +17,16 @@ public class _168 {
     public String convertToTitle_accepted_more_beautiful(int n) {
         /**Get the right most digit first, move to the left, e.g. when n = 28, we get 'B' first, then we get 'A'.*/
         StringBuilder sb = new StringBuilder();
-        while(n != 0){
-            int temp = (n-1)%26;
-            sb.append((char)(temp+65));
-            n = (n-1)/26;
+        while (n != 0) {
+            int temp = (n - 1) % 26;
+            sb.append((char) (temp + 65));
+            n = (n - 1) / 26;
         }
         return sb.reverse().toString();
-    
+
     }
-    
-    public static void main(String...strings){
+
+    public static void main(String... strings) {
         _168 test = new _168();
 //        int n = 28899;
 //        int n = 1;
@@ -51,13 +51,13 @@ public String convertToTitle_accepted(int n) {
         /**'Z' is the corner case, so we'll have to special case handling specially, also, we'll have to do (n-1)/26,
          * only when this is not equal to 1, we'll continue.*/
         StringBuilder sb = new StringBuilder();
-        while(n != 0){
-            int temp = n%26;
-            if(temp == 0) sb.append("Z");
-            else sb.append((char)(temp+64));
-            n = (n-1)/26;
+        while (n != 0) {
+            int temp = n % 26;
+            if (temp == 0) sb.append("Z");
+            else sb.append((char) (temp + 64));
+            n = (n - 1) / 26;
         }
         return sb.reverse().toString();
     }
-    
+
 }

src/main/java/com/fishercoder/solutions/_169.java

@@ -12,59 +12,59 @@
 
 */
 public class _169 {
-    
-    public int majorityElement_bit_manipulation(int[] nums){
+
+    public int majorityElement_bit_manipulation(int[] nums) {
         int[] bit = new int[32];//because an integer is 32 bits, so we use an array of 32 long
-        for(int num : nums){
-            for(int i = 0; i < 32; i++){
-                if((num >> (31-i) & 1) == 1) bit[i]++;//this is to compute each number's ones frequency
+        for (int num : nums) {
+            for (int i = 0; i < 32; i++) {
+                if ((num >> (31 - i) & 1) == 1) bit[i]++;//this is to compute each number's ones frequency
             }
         }
         int res = 0;
         //this below for loop is to construct the majority element: since every bit of this element would have appeared more than n/2 times
-        for(int i = 0; i < 32; i++){
-            bit[i] = bit[i] > nums.length/2 ? 1 : 0;//we get rid of those that bits that are not part of the majority number
-            res += bit[i]*(1 << (31-i));
+        for (int i = 0; i < 32; i++) {
+            bit[i] = bit[i] > nums.length / 2 ? 1 : 0;//we get rid of those that bits that are not part of the majority number
+            res += bit[i] * (1 << (31 - i));
         }
         return res;
     }
-    
+
     //saw a really clever solution on Discuss, though it didn't use bit manipulatoin
     //this is actually applying a famous algorithm called Moore Voting algorithm: http://www.cs.utexas.edu/~moore/best-ideas/mjrty/example.html
-    public int majorityElement_moore_voting_algorithm(int[] nums){
+    public int majorityElement_moore_voting_algorithm(int[] nums) {
         int count = 1, majority = nums[0];
-        for(int i = 1; i < nums.length; i++){
-            if(count == 0){
+        for (int i = 1; i < nums.length; i++) {
+            if (count == 0) {
                 count++;
                 majority = nums[i];
-            } else if(nums[i] == majority){
+            } else if (nums[i] == majority) {
                 count++;
             } else count--;
         }
         return majority;
     }
-    
-    public static void main(String...strings){
-        int[] nums = new int[]{1,2,3,4,2,3,2,2,4,2};
+
+    public static void main(String... strings) {
+        int[] nums = new int[]{1, 2, 3, 4, 2, 3, 2, 2, 4, 2};
         _169 test = new _169();
         System.out.println(test.majorityElement_bit_manipulation(nums));
     }
-    
+
     //my natural idea is to either compute the frequency of each unique number or sort it and return the median, I can hardly think of 
     //how bit manipulation could come into play for this question
     //this is O(n) time.
     public int majorityElement_compute_frequency(int[] nums) {
         Map<Integer, Integer> map = new HashMap();
-        for(int i : nums){
+        for (int i : nums) {
             map.put(i, map.getOrDefault(i, 0) + 1);
-            if(map.get(i) > nums.length/2) return i;
+            if (map.get(i) > nums.length / 2) return i;
         }
         return -1;
     }
-    
+
     //This is O(nlogn) time.
     public int majorityElement_sort(int[] nums) {
         Arrays.sort(nums);
-        return nums[nums.length/2];
+        return nums[nums.length / 2];
     }
 }

src/main/java/com/fishercoder/solutions/_17.java

@@ -21,23 +21,24 @@ public class _17 {
 
     public List<String> letterCombinations(String digits) {
         List<String> result = new ArrayList();
-        if(digits.length() == 0) return result;
+        if (digits.length() == 0) return result;
 
         String[] digits2Letters = new String[]{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
 
         result.add("");//this line is important, otherwise result is empty and Java will default it to an empty String
-        for(int i = 0; i < digits.length(); i++){
-            result = combine(digits2Letters[digits.charAt(i)-'0'], result);
+        for (int i = 0; i < digits.length(); i++) {
+            result = combine(digits2Letters[digits.charAt(i) - '0'], result);
         }
 
         return result;
     }
 
-    List<String> combine(String letters, List<String> result){
+    List<String> combine(String letters, List<String> result) {
         List<String> newResult = new ArrayList();
 
-        for(int i = 0; i < letters.length(); i++){//the order of the two for loops doesn't matter, you could swap them and it still works.
-            for(String str : result){
+        for (int i = 0; i < letters.length(); i++) {
+            //the order of the two for loops doesn't matter, you could swap them and it still works.
+            for (String str : result) {
                 newResult.add(str + letters.charAt(i));
             }
         }

src/main/java/com/fishercoder/solutions/_171.java

@@ -17,10 +17,10 @@ public class _171 {
     public int titleToNumber(String s) {
         char[] c = s.toCharArray();
         int result = 0;
-        for(int i = s.length()-1; i >= 0; i--){
-            result += (c[i]-64) * ((int) Math.pow(26, s.length()-i-1));//The ASCII value of A is 65
+        for (int i = s.length() - 1; i >= 0; i--) {
+            result += (c[i] - 64) * ((int) Math.pow(26, s.length() - i - 1));//The ASCII value of A is 65
         }
         return result;
     }
-    
+
 }

src/main/java/com/fishercoder/solutions/_172.java

@@ -6,7 +6,7 @@ public class _172 {
 
     public int trailingZeroes(int n) {
         int result = 0;
-        while(n > 4){
+        while (n > 4) {
             n /= 5;
             result += n;
         }