标签:tee sum cte number racket its any span form
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".
这道题感觉挺难的,迭代的思想感觉不太容易想
class Solution { public String decodeString(String s) { if(s == null || s.length() == 0){ return ""; } Stack<Integer> numStack = new Stack<>(); Stack<String> strStack = new Stack<>(); int count = 0; StringBuilder res = new StringBuilder(); for(int i =0; i< s.length(); i++){ char c = s.charAt(i); //find number if(c >= ‘0‘ && c <= ‘9‘){ count = 10 * count + c - ‘0‘; } //find left square brackets else if(c == ‘[‘){ numStack.push(count); strStack.push(res.toString()); count = 0; res = new StringBuilder(); } //find right square brackets else if( c == ‘]‘){ StringBuilder temp = new StringBuilder(strStack.pop()); int repeat = numStack.pop(); for(int j = 0; j < repeat; j++){ temp.append(res); } res = temp; } //find normal character else{ res.append(c); } } return res.toString(); } }
标签:tee sum cte number racket its any span form
原文地址:https://www.cnblogs.com/incrediblechangshuo/p/9733713.html