class Solution { private int i = -1;//全局变量i,记录字符数组指针位置 public String decodeString(String s) { return dfs(s.toCharArray(), s.length()).toString(); } //递归函数 private StringBuilder dfs(char[] chars, int len) { int num = 0; StringBuilder str = new StringBuilder(); wh...
Given an encoded string, return its decoded string. The encoding rule is:k[encoded_string], where theencoded_stringinside the square brackets is being repeated exactlyktimes. Note thatkis guaranteed to be a positive integer. You may assume that the input string is always valid; there are no ...
publicclassSolution {publicString decodeString(String s) { StringBuilder builder=newStringBuilder(); decodeStringRecur(s.toCharArray(),builder,0);returnbuilder.toString(); }publicintdecodeStringRecur(char[] sArr, StringBuilder builder,intstart){if(start>=sArr.length){returnstart; }intp1 =start;while...
C++ 智能模式 1 2 3 4 5 6 class Solution { public: string decodeString(string s) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2Case 3 s = "3[a]2[bc]" 1 2 3 "3[a]2[bc]" "3[a2[c]]" "2[abc]3[cd]ef" Source ...
classSolution {public:stringdecodeString(strings) {stringt; stack<int>s_num; stack<string>s_str;intcnt =0;for(inti =0; i < s.size(); ++i) {if(s[i] >='0'&& s[i] <='9') { cnt=10* cnt + s[i] -'0'; }elseif(s[i] =='[') { ...
Can you solve this real interview question? Decode String - Given an encoded string, return its 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 i
链接:https://leetcode-cn.com/problems/decode-string 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 二,解题思路 核心问题:寻找匹配的左右括号(找到与最左侧括号相匹配的右括号,即可对中间的字符串递归调用算法)。
Leetcode: Decode String Given an encoded string,returnit'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....
leetcode394. Decode String 代码语言:javascript 代码运行次数:0 运行 AI代码解释 Given an encoded string,returnit'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 ...
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