refactor for format

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_11.java

@@ -10,7 +10,8 @@ public class _11 {
     public int maxArea(int[] height) {
         int max = Integer.MIN_VALUE;
         int len = height.length;
-        int i = 0, j = len - 1;
+        int i = 0;
+        int j = len - 1;
         while (i < j) {
             if (Math.min(height[i], height[j]) * (j - i) > max) {
                 max = Math.min(height[i], height[j]) * (j - i);

src/main/java/com/fishercoder/solutions/_115.java

@@ -11,7 +11,8 @@ public class _115 {
      * 4. starting from (1,1)*/
 
     public int numDistinct(String s, String t) {
-        int m = s.length(), n = t.length();
+        int m = s.length();
+        int n = t.length();
         int[][] dp = new int[m + 1][n + 1];
 
         char[] schar = s.toCharArray();

src/main/java/com/fishercoder/solutions/_123.java

@@ -13,8 +13,10 @@ public class _123 {
     //this is a very clear solution and very highly upvoted in Discuss, but not extensibel to K solution.
     public int maxProfit(int[] prices) {
         if (prices.length < 2) return 0;
-        int buy1 = Integer.MIN_VALUE, buy2 = Integer.MIN_VALUE;//we use negative numbers to denote buy1 and buy2, thus use Integer.MIN_VALUE here is more convenient.
-        int sell1 = 0, sell2 = 0;
+        int buy1 = Integer.MIN_VALUE;
+        int buy2 = Integer.MIN_VALUE;//we use negative numbers to denote buy1 and buy2, thus use Integer.MIN_VALUE here is more convenient.
+        int sell1 = 0;
+        int sell2 = 0;
         for (int i = 0; i < prices.length; i++) {
             buy1 = Math.max(buy1, -prices[i]);
             sell1 = Math.max(sell1, buy1 + prices[i]);