标签：remember mask please reset otherwise ret public tor pad
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 ‘ ‘
思路：
需要注意的是，数字可能是大于10的也就是说不一定是一位数。。。
参考了大神的代码。。
class StringIterator { string a; size_t i = 0, c = 0; char ch; public: StringIterator(string a) { this->a = a; } char next() { if (c) return c--, ch; if (i >= a.size()) return ‘ ‘; ch = a[i++]; while (i < a.size() && isdigit(a[i])) c = c*10+a[i++]-‘0‘; c--; return ch; } bool hasNext() { return c || i < a.size(); } };
参考：
[leetcode-604-Design Compressed String Iterator]
标签：remember mask please reset otherwise ret public tor pad
原文地址：http://www.cnblogs.com/hellowooorld/p/6984798.html
