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Given an encoded string, return it‘s decoded string.
The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.
You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.
Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won‘t be input like 3a or 2[4].
Examples:
s = "3[a]2[bc]", return "aaabcbc". s = "3[a2[c]]", return "accaccacc". s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
public class Solution { public String decodeString(String s) { StringBuilder builder = new StringBuilder(); decodeStringRecur(s.toCharArray(),builder,0); return builder.toString(); } public int decodeStringRecur(char[] sArr, StringBuilder builder, int start){ if (start>=sArr.length){ return start; } int p1 = start; while (p1<sArr.length && sArr[p1]!=‘]‘){ if (sArr[p1]<‘0‘ || sArr[p1]>‘9‘){ builder.append(sArr[p1++]); } else { // get the following encoded string. // get the number first. int val = 0; while (p1<sArr.length && sArr[p1]!=‘[‘){ val = val*10 + (int)(sArr[p1++]-‘0‘); } // get the string. StringBuilder subBuilder = new StringBuilder(); p1 = decodeStringRecur(sArr,subBuilder,p1+1); // add into decoded string. for (int i=0;i<val;i++){ builder.append(subBuilder); } } } return (p1<sArr.length) ? p1+1 : p1; } }
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原文地址：http://www.cnblogs.com/lishiblog/p/5874147.html
