牛客NC199 字符串解码【中等递归，栈的思想 C++/Java/Go/PHP】

题目

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题目链接：
https://www.nowcoder.com/practice/4e008fd863bb4681b54fb438bb859b92
相同题目：
https://www.lintcode.com/problem/575

思路

解法和基础计算器1，2,3类似,递归

参考答案C++

struct Info {string str;int stopindex;Info(string e, int c) : str(e), stopindex(c) {}
};class Solution {public:/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param s string字符串* @return string字符串*/string decodeString(string s) {//解法和基础计算器1，2,3类似,递归return process(s, 0).str;}Info process(string s, int idx) {string str;int cur = 0;while (idx < s.size() && s[idx] != ']') {char c = s[idx];if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')) {str += c;idx++;} else if (c >= '0' && c <= '9') {cur = cur * 10 + int(c - '0');idx++;} else { //遇到[  需要递归获取[...] 的结果Info info = process(s, idx + 1);str += getStr(info.str, cur);cur = 0;idx = info.stopindex + 1;}}return Info(str, idx);}string getStr(string s1, int count) {string ans;for (int i = 0; i < count; i++) {ans += s1;}return ans;}
};

参考答案Java

import java.util.*;public class Solution {/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param s string字符串* @return string字符串*/public String decodeString (String s) {//解法和基础计算器1，2,3类似,递归return process(s, 0).ans;}public Info process(String s, int i) {StringBuilder ans = new StringBuilder();int cur = 0;while (i < s.length() && s.charAt(i) != ']') {char c = s.charAt(i);if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')) {ans.append(c);i++;} else if (c >= '0' && c <= '9') {cur = cur * 10 + c - '0';i++;} else { //遇到[Info next = process(s, i + 1);ans.append(getStr(next.ans, cur));cur = 0;i = next.stopIndex + 1;}}return new Info(ans.toString(), i);}public String getStr(String str, int count) {StringBuilder sb = new StringBuilder();for (int i = 0; i < count ; i++) {sb.append(str);}return sb.toString();}static class Info {String ans;int stopIndex;public Info(String s, int e) {ans = s;stopIndex = e;}}
}

参考答案Go

package mainimport "bytes"/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param s string字符串* @return string字符串*/
func decodeString(s string) string {//本答案和解答基础计算器1,2,3的思路类似//递归。遇到字母，遇到数字，遇到[  三种情况return process(s, 0).str
}func process(s string, idx int) Info {var buf bytes.Buffercur := 0for idx < len(s) && s[idx] != ']' {c := s[idx]if (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') {buf.WriteString(string(c))idx++} else if c >= '0' && c <= '9' {cur = cur*10 + int(c-'0')idx++} else { //遇到[  递归去吧，我需要递归的结果info := process(s, idx+1)buf.WriteString(getStr(info.str, cur))cur = 0idx = info.stopindex + 1}}return Info{buf.String(), idx}
}func getStr(str string, count int) string {var buf bytes.Bufferfor i := 0; i < count; i++ {buf.WriteString(str)}return buf.String()
}type Info struct {str       stringstopindex int
}

参考答案PHP

<?php/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可** * @param s string字符串 * @return string字符串*/
function decodeString( $s )
{//本答案和解答基础计算器1,2,3的思路类似//递归。遇到字母，遇到数字，遇到[  三种情况return process($s,0)->str;
}function process($s,$idx) {$str ='';$cur=0;while ($idx <strlen($s) && $s[$idx] !=']') {$c = $s[$idx];if(($c>='a' && $c<='z') ||($c>='A' && $c <='Z')){$str.=$c;$idx++;}else if($c>='0' && $c<='9'){$cur =$cur*10+intval($c-'0');$idx++;}else{ //遇到[ 了，递归获取[...] 中等的结果$info = process($s,$idx+1);$str.=(getStr($info->str,$cur));$cur =0;$idx = $info->stopindex+1;}}return new Info($str,$idx);
}function getStr($s,$count){$str = '';for($i=0;$i<$count;$i++){$str.=$s;}return $str;
}class Info{public $str;public $stopindex;public function __construct($a,$b){$this->str = $a;$this->stopindex = $b;}
}