refactor 97

fishercoder1534 · fishercoder1534 · commit 945bcfd78e6e · 2020-05-09T10:12:22.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_97.java b/src/main/java/com/fishercoder/solutions/_97.java
@@ -1,57 +1,46 @@
 package com.fishercoder.solutions;
 
-/**
- * 97. Interleaving String
- *
- * Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
- * For example,
- * Given:
- * s1 = "aabcc",
- * s2 = "dbbca",
- * When s3 = "aadbbcbcac", return true.
- * When s3 = "aadbbbaccc", return false.
- */
 public class _97 {
-  public static class Solution1 {
-    public boolean isInterleave(String s1, String s2, String s3) {
-      int m = s1.length();
-      int n = s2.length();
-      if (m + n != s3.length()) {
-        return false;
-      }
+    public static class Solution1 {
+        public boolean isInterleave(String s1, String s2, String s3) {
+            int m = s1.length();
+            int n = s2.length();
+            if (m + n != s3.length()) {
+                return false;
+            }
 
-      boolean[][] dp = new boolean[m + 1][n + 1];
+            boolean[][] dp = new boolean[m + 1][n + 1];
 
-      dp[0][0] = true;
+            dp[0][0] = true;
 
-      for (int i = 0; i < m; i++) {
-        if (s1.charAt(i) == s3.charAt(i)) {
-          dp[i + 1][0] = true;
-        } else {
-          //if one char fails, that means it breaks, the rest of the chars won't matter any more.
-          //Mian and I found one missing test case on Lintcode: ["b", "aabccc", "aabbbcb"]
-          //if we don't break, here, Lintcode could still accept this code, but Leetcode fails it.
-          break;
-        }
-      }
+            for (int i = 0; i < m; i++) {
+                if (s1.charAt(i) == s3.charAt(i)) {
+                    dp[i + 1][0] = true;
+                } else {
+                    //if one char fails, that means it breaks, the rest of the chars won't matter any more.
+                    //Mian and I found one missing test case on Lintcode: ["b", "aabccc", "aabbbcb"]
+                    //if we don't break, here, Lintcode could still accept this code, but Leetcode fails it.
+                    break;
+                }
+            }
 
-      for (int j = 0; j < n; j++) {
-        if (s2.charAt(j) == s3.charAt(j)) {
-          dp[0][j + 1] = true;
-        } else {
-          break;
-        }
-      }
+            for (int j = 0; j < n; j++) {
+                if (s2.charAt(j) == s3.charAt(j)) {
+                    dp[0][j + 1] = true;
+                } else {
+                    break;
+                }
+            }
 
-      for (int i = 1; i <= m; i++) {
-        for (int j = 1; j <= n; j++) {
-          int k = i + j - 1;
-          dp[i][j] = (s1.charAt(i - 1) == s3.charAt(k) && dp[i - 1][j])
-              || (s2.charAt(j - 1) == s3.charAt(k) && dp[i][j - 1]);
-        }
-      }
+            for (int i = 1; i <= m; i++) {
+                for (int j = 1; j <= n; j++) {
+                    int k = i + j - 1;
+                    dp[i][j] = (s1.charAt(i - 1) == s3.charAt(k) && dp[i - 1][j])
+                            || (s2.charAt(j - 1) == s3.charAt(k) && dp[i][j - 1]);
+                }
+            }
 
-      return dp[m][n];
+            return dp[m][n];
+        }
     }
-  }
 }
