LeetCode 394. 字符串解码

 

思路#

方法：用栈模拟#

class Solution {
public:
    string decodeString(string s) {
        deque<char> d;

        for(int i = 0; i < s.length(); ++i) {
            if(s[i] == ']') {
                deque<char> tmp;
                while(d.back() != '[') {
                    tmp.push_front(d.back());
                    d.pop_back();
                }

                d.pop_back();  //左括号'['出栈
                //获取循环次数repeat，注意repeat可能大于1位数
                int repeat = 0, r = 1;
                while(!d.empty() && isdigit(d.back())) {
                    repeat += r * (d.back() - '0');
                    r *= 10;
                    d.pop_back();
                }
                
                while(repeat--) {
                    for(int j = 0; j < tmp.size(); ++j) {
                        d.push_back(tmp[j]);
                    }
                }
            } else {
                d.push_back(s[i]);
            } 
        }

        string res(d.begin(), d.end());

        return res;
    }
};

复杂度分析：#

时间复杂度：O(S)，S为答案字符串的长度

空间复杂度：O(S)