leetcode 394. Decode String 字符串解码 + DFS深度优先遍历求解

ゝ一纸荒年。 2022-06-04 06:23 196阅读 0赞

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”.

这道题十分简单，就是做一个遍历即可，按照规则直接展开即可

**第一反应想到的方法就是直接按照括号去做展开接口，注意提取数字和括号内的东西，就是直接处理字符串，有点麻烦，其余的就没什么了**

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    
    class Solution 
    {
    public:
        string decodeString(string s) 
        {
            return dfs(s);
        }
    
        string dfs(string s)
        {
            int start = s.find_first_of('[');
            if (start == s.npos)
                return s;
            else
            {
                int i = start,count = 0;
                for (; i < s.length(); i++)
                {
                    if (s[i] == '[')
                        count++;
                    else if (s[i] == ']')
                        count--;
                    if (count==0)
                        break;
                }
    
                string left = s.substr(0, start); 
                string cen = s.substr(start + 1, i - start - 1);
                string right = s.substr(i + 1);
    
                for (i = left.length() - 1; i >= 0; i--)
                {
                    if (!(left[i]>='0' && left[i]<='9'))
                        break;
                }
    
                int size = stoi(left.substr(i + 1));
                left = left.substr(0, i + 1);
                for (int j = 0; j < size; j++)
                    left += cen;
    
                return dfs(left + right);
            }
        }
    };