LeetCode-394. Decode String

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Description

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].

Example

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

Solution 1(C++)

class Solution {
public:
    string decodeString(string s) {
        using P = pair<int,string>; 
        stack<P> st;                        // stack stores pair of {k,encoded_string}
        int k = 0;                          // k-value
        string res = "";                    // result
        for (const auto& c: s) {
            if (isdigit(c))             // if char is digit, then evaluate k-value
                k = k * 10 + c - '0';
            else if (c == '[') {          // push k, and a placeholder string on stack
                st.push({k,""});
                k = 0;          
            } else if (c == ']') {                  // found ']', get stack top element
                P top = st.top(); st.pop();
                while (top.first-- > 0) {
                    if (st.empty())              // stack empty means, '[' ']' non-nested case, so add to result directly
                        res += top.second;
                    else 
                        st.top().second += top.second;   // nested case, so append to next stack entry
                }
            } else {
                if (!st.empty())    
                    st.top().second += c;
                else
                    res += c;                   // 2[abc]3[cd]ef: to handle non-encoded chars (ef), add to result directly 
            }
        }

        while(!st.empty()) {                // populate result if stack not empty (nested case)
            P top = st.top(); st.pop();
            while (top.first-- > 0)
                res += top.second;
        }
        return res;
    }
};

Solution 2(C++)

class Solution {
public:
    string decodeString(string s) {
        int index = 0;
        return helper(s, index);
    }

private:
    string helper(string s, int& index){
        string res;

        while(index < s.size() && s[index] != ']'){
            if(!isdigit(s[index])){
                res += s[index++];
            }
            else{
                int times = 0;
                while(isdigit(s[index])){
                    times = times*10 + (s[index++]-'0');
                }

                index++;
                string temp = helper(s, index);
                index++;

                while(times-- > 0){
                    res += temp;
                }
            }
        }

        return res;
    }
};

算法分析

这道题要说的很详细也没有必要，主要还是编程能力的体现，第一遍我是没做出来。学习这道题的解法推荐解法二，简单易懂，利用了递归的解法。

程序分析

略。