LeetCode Design Compressed String Iterator
Design and implement a data structure for a compressed string iterator. It should support the following operations: next and hasNext.
The given compressed string will be in the form of each letter followed by a positive integer representing the number of this letter existing in the original uncompressed string.
next() – if the original string still has uncompressed characters, return the next letter; Otherwise return a white space.
hasNext() – Judge whether there is any letter needs to be uncompressed.
Note:
Please remember to RESET your class variables declared in StringIterator, as static/class variables are persisted across multiple test cases. Please see here for more details.
Example:
StringIterator iterator = new StringIterator("L1e2t1C1o1d1e1");
iterator.next(); // return 'L'
iterator.next(); // return 'e'
iterator.next(); // return 'e'
iterator.next(); // return 't'
iterator.next(); // return 'C'
iterator.next(); // return 'o'
iterator.next(); // return 'd'
iterator.hasNext(); // return true
iterator.next(); // return 'e'
iterator.hasNext(); // return false
iterator.next(); // return ' '
本题设计了一种压缩字符串格式，即把多个连续相同字符用单个字符已经频率代替了，比如”L1e2t1C1o1d1e1″展开其实就是”LeetCode”。现在要针对这种压缩字符串，设计一个迭代器，实现next和hasNext函数。 我一开始上暴力，即首先把压缩字符串展开成常规字符串，然后直接下标操作。代码如下： [cpp] class StringIterator { private: string line; int cur, n; public: StringIterator(string compressedString) { line = ""; cur = n = 0; vector<char> letters; vector<int> rep; string digits = ""; for (int i = 0; i < compressedString.size(); ++i) { if (isalpha(compressedString[i])) { letters.push_back(compressedString[i]); if (digits != "") { rep.push_back(atoi(digits.c_str())); digits = ""; } } else digits += string(1, compressedString[i]); } rep.push_back(atoi(digits.c_str())); for (int i = 0; i < letters.size(); ++i) { line += string(rep[i], letters[i]); } n = line.size(); } char next() { if (cur < n)return line[cur++]; else return ‘ ‘; } bool hasNext() { return cur < n; } }; [/cpp] 本代码MLE了。因为有个样例是“a1234567890b1234567890”，我表示无语，如果展开确实会MLE。 其实写上面的代码的时候我就知道可以优化，不对压缩字符串展开，而是表示成(a,1234567890)和(b,1234567890)，next的时候只对频率递减。这样只需要存储少量字符以及对应的频率。 代码如下： [cpp] typedef long long ll; class StringIterator { private: int cur, n; vector<char> letters; vector<ll> rep; public: StringIterator(string compressedString) { cur = n = 0; string digits = ""; for (int i = 0; i < compressedString.size(); ++i) { if (isalpha(compressedString[i])) { letters.push_back(compressedString[i]); if (digits != "") { rep.push_back(atoll(digits.c_str())); digits = ""; } } else digits += string(1, compressedString[i]); } rep.push_back(atoi(digits.c_str())); n = letters.size(); } char next() { if (rep[cur] > 0) { char ans = letters[cur]; –rep[cur]; return ans; } else { if (cur == n – 1)return ‘ ‘; else { char ans = letters[++cur]; –rep[cur]; return ans; } } } bool hasNext() { return cur < n – 1 || (cur == n – 1 && rep[cur] > 0); } }; [/cpp] 本代码提交AC，用时13MS。]]>