solution for lc686

Author: maslke

String.md

@@ -3,6 +3,7 @@ No.|Title|Difficulty|Solved|Date
 --|:--:|--:|--:|--:|
 58|[Length of Last Word](https://leetcode.com/problems/length-of-last-word/)|Easy|yes|2019-01-08
 680|[Valid Palindrome II](https://leetcode.com/problems/valid-palindrome-ii/)|Easy|yes|2019-1-14
+686|[Repeated String Match](https://leetcode.com/problems/repeated-string-match/)|Easy|yes|2019-1-14
 696|[Count Binary Substrings](https://leetcode.com/problems/count-binary-substrings/)|Easy|yes|2019-01-11
 848|[Shifting Letters](https://leetcode.com/problems/shifting-letters/)|Medium|yes|2019-01-12
 859|[Buddy Strings](https://leetcode.com/problems/buddy-strings/)|Easy|yes|2019-01-09
@@ -47,6 +48,10 @@ public boolean validPalindrome(String s) {
     }
 ```
 
+686. [Repeated String Match](https://leetcode.com/problems/repeated-string-match/)
+
+字符串问题。计算字符串A叠加多少次之后，可以包含字符串B。注意退出条件即可。
+
 696. [Count Binary Substrings](https://leetcode.com/problems/count-binary-substrings/)
 
 对于给定的字符串，可以形成 m * 1 + n * 0 + m1 * 1 + n1 * n这种形式。

src/main/java/easy/lc686.java

@@ -0,0 +1,24 @@
+//https://leetcode.com/problems/repeated-string-match/
+//686. Repeated String Match
+class Solution {
+    public int repeatedStringMatch(String A, String B) {
+        StringBuilder sb = new StringBuilder(A);
+        int count = 1;
+        while (true) {
+            if (sb.length() < B.length()) {
+                sb.append(A);
+                count++;
+            } else {
+                if (sb.indexOf(B) >= 0) {
+                    return count;
+                } else {
+                    sb.append(A);
+                    count++;
+                    if (sb.indexOf(B) >= 0) return count;
+                    return -1;
+                }
+            }
+        }
+       // return -1;
+    }
+}