public String decodeString(String s) { result += e.getStr(); public Exp(char ch) { else if(s.charAt(i)=='['){ Stack temp =new Stack(); Stack str =new Stack(); while( !temp.isEmpty()){ str.push(str.pop()+s.charAt(i)); Exp top = stack.pop(); if (p == '0') { if (c == '0') { c = ch; String getMultipliedString(String x,int n){ while( !temp.isEmpty()) An empty digit sequence is considered to have one decoding. int count; It is similar to the problem of counting ways of climbing stairs. It may be assumed that the input contains valid digits from 0 to 9 and there are no leading 0’s, no … char c; dp[i + 1] = dp[i] + dp[i - 1]; The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. } Machine 1 (sender) has the function: string encode (vector strs) { //... your code return encoded_string; } 0. praveen0989 12. stack.peek().list.add(new Exp(c)); i++; as per base64 encoding algorithm and returns and encoded byte array, which can be converted into String. This is a really interesting problem. Java | Runtime: 0 ms, faster than 100.00% | Memory, less than 99.53% | With Comments The key to solve this problem is convert the string to a structured data structure and recursively form the return string. }, String sf=""; "[".equals(str.peek()) ){ num = num + c; Exp e = new Exp(1); } Contribute to zhangyu345293721/leetcode development by creating an account on GitHub. int value = Integer.parseInt(num); return 0; https://www.programcreek.com/2014/09/leetcode-decode-string-java/ } for (int i = 0; i < s.length(); i++) { if(!num.equals("")){ The encoded string is then sent over the network and is decoded back to the original list of strings. } else if (c == '[') { return dp[s.length()]; result += c; if( !str.isEmpty()&& i> Strings >> Decode Ways Leetcode Java. Stack nums=new Stack(); Huifeng Guan 141 views. }, Solution Using two stacks. String result = ""; Raw. length() … LeetCode – Decode String (Java) - Program Creek. return sf; if (c == '0') { while(i> Algorithms If you want someone to read your code, please put the code inside
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tags. dp[i + 1] = dp[i]; } else { public String decodeString(String s) { Back To Back SWE 25,552 views stack.push(exp); int[] dp = new int[s.length() + 1]; The relation is dp[n]=dp[n-1]+dp[n-2]. i++; A message containing letters from A-Z is being encoded to numbers using the following mapping: 'A' -> 1 dp[0] = 1; For example, given s = "leetcode", dict = ["leet", "code"]. sf+=temp.pop(); } public String decodeString (String s) { if (s == null || s.length() == 0) { return s; } Stack countStack = new Stack<>(); Stack< String > resultStack = new Stack<>(); char[] strArr = s.toCharArray(); int count = 0; String curResult = ""; for (int i = 0; i < s.length(); i++) { //calculate repeat number if (Character.isDigit(strArr[i])) { count = count * 10 + (strArr[i] - '0'); } //push previous decoded … Decode Ways Leetcode Java. return result; } } ... LeetCode – One Edit Distance (Java) LeetCode – Isomorphic Strings (Java) Category >> Algorithms If you want someone to read your code, please put the code inside
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tags. stack.push(e); January 7, 2016 3:14 pm | Leave a Comment | crazyadmin. For … } str.push(""); return res; } dp[i + 1] = dp[i - 1]; Exp root = stack.pop(); You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc. Example 1: Input: s = "12" Output: 2 Explanation: "12" could be decoded as "AB" (1 2) or "L" (12). } Decode String, 08/18/2019 - Duration: 11:26. } The answer is guaranteed to fit in a 32-bit integer. dp[1] = 1; Given a non-empty string num containing only digits, return the number of ways to decode it. JAVA RECURSIVE CODE USING STACK 1 PASS 0MS 100% FAST SOLUTION. temp.push(str.pop()); } } This is one of Facebook's favorite interview questions to ask! public Exp(int num) { res+=x; Java Solution. if (num.length() == 0) Return true because "leetcode" can be segmented as "leet code". while(n-->0){ We will see that it is not that complicated when using recursion, and that the problem can be … temp.push(str.pop()); There are some cases to } 30 VIEWS. For example: i++; dependency in pom.xml: } else if (c == ']') { } if (s.charAt(0) == '0') { stack.peek().list.add(top); str.push(getMultipliedString(sf,nums.pop()) ); Problem: A message containing letters from A-Z is being encoded to numbers using the following mapping: ‘A’ – 1 ‘B’ – 2 It is similar to the problem of counting ways of climbing stairs. dp[i + 1] = dp[i - 1]; numDecodingEndingHere[index+1] += numDecodingEndingHere[index] # We can also decode current character and the next one at the smae # time, and then move to the second next position. 11:26. String sf=""; } } if (c == '0' && (p == '0' || p > '2')) { 'B' -> 2 https://leetcode.com/problems/decode-string/eval(ez_write_tag([[728,90],'programcreek_com-medrectangle-3','ezslot_4',136,'0','0'])); The key to solve this problem is convert the string to a structured data structure and recursively form the return string.eval(ez_write_tag([[580,400],'programcreek_com-medrectangle-4','ezslot_2',137,'0','0'])); class Solution { Java || Stack || Modular code. Stack stack = new Stack<>(); EncodeandDecodeStrings.java. Encoding & Decoding String into Base64 Java. 37 VIEWS. } public static StringBuilder decode (String s) { // Stack is deprecated so using double-ended Q Deque multipliers = new ArrayDeque<>(); Deque result = new ArrayDeque<>(); result.push(new StringBuilder()); int multiplier = 0; // Would be nice to use an 'enhanced' for loop, but don't want // the expense of converting the String to an array (ie toCharArray) // for (char ch : … Java Solution. https://youtu.be/Km4iqih6WjI, LeetCode – Letter Combinations of a Phone Number (Java). else{ Leetcode solutions. 1. Example 2: Input: S = "ha22", K = 5 Output: "h" Explanation: The decoded string is "hahahaha". # We can decode current character and move to the next position. Stack temp =new Stack(); public String getStr() { char p = s.charAt(i - 1); 0. shivam_gupta_ 1. char c = s.charAt(i); dp[i + 1] = dp[i]; for (int i = 0; i < count; i++) { Total Ways To Decode A String - Recursive Dynamic Programming Approach ("Decode Ways" on LeetCode) - Duration: 11:58. 'Z' -> 26eval(ez_write_tag([[250,250],'programcreek_com-medrectangle-3','ezslot_4',136,'0','0'])); Given an encoded message containing digits, determine the total number of ways to decode it. while( !str.isEmpty()&& ! 249 LeetCode Java: Group Shifted Strings – Easy 250 LeetCode Java: Count Univalue Subtrees – Medium Solutions 251 - 300 251 Flatten 2D Vector 252 LeetCode Java: Meeting Rooms – Easy ... Encode and Decode Strings Problem: Design an algorithm to encode a list of strings to a string. Exp exp = new Exp(value); LeetCode - Encode and Decode Strings. Given an encoded string, return its decoded string. String x= str.pop(); class Solution { public int[] decode(int[] encoded, int first) { int decode [] = new int[encoded.length+1]; decode[0] = first; for(int i=1;i= '0' && c <= '9') { } List list; Example 2: num = ""; Note that k is guaranteed to be a positive integer. } nums.push(Integer.valueOf(num)); Example 3: int i=0; return root.getStr(); if (list != null) { The encoded string is then sent over the network and is decoded back to the original list of strings. In order to execute this Base64 Encoding Example in Java, you need to download and add commons-codec-1.2.jar into your application classpath. The relation is dp[n]=dp[n-1]+dp[n-2]. LeetCode – Multiply Strings (Java) LeetCode – ZigZag Conversion (Java) LeetCode – Serialize and Deserialize Binary Tree (Java) Category >> Algorithms >> Interview >> Java If you want someone to read your code, please put the code inside
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tags. This problem can be solve by using dynamic programming. } } else { // Design an algorithm to encode a list of strings to a string. count = num; } } while(i= '0' && c <= '9') { num = num + c; } else if ( c == ' [') { if ( num. } else if (c <= '6') { for (Exp e : list) { } Naive Approach { class Solution { public String decodeString (String s) { Stack < Exp > stack = new Stack <>(); Exp e = new Exp (1); stack. }, LeetCode – Longest Valid Parentheses (Java). str.push("["); } For example, given "3[a2[b]]", return "abbabbabb". list = new ArrayList<>(); else if(s.charAt(i)==']'){ Leetcode刷题之旅. Number of ways to Decode it Duration: 11:26, 08/18/2019 - Duration: 11:26 Java RECURSIVE code using 1! Climbing stairs number of ways to Decode it the key to solve this can. | crazyadmin is: k [ encoded_string ], where the encoded_string the..., dict = [ `` leet '', return `` abbabbabb '' Facebook 's favorite interview questions to ask,. 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