604. Design Compressed String Iterator 迭代压缩字符串

• 作者： 负雪明烛
• id： fuxuemingzhu
• 个人博客：https://fuxuemingzhu.cn/

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题目地址：https://leetcode-cn.com/problems/design-compressed-string-iterator/

题目描述

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 ' '

题目大意

对于一个压缩字符串，设计一个数据结构，它支持如下两种操作： next 和 hasNext。 给定的压缩字符串格式为：每个字母后面紧跟一个正整数，这个整数表示该字母在解压后的字符串里连续出现的次数。

• next() - 如果压缩字符串仍然有字母未被解压，则返回下一个字母，否则返回一个空格。
• hasNext() - 判断是否还有字母仍然没被解压。

解题方法

维护当前字符和次数

这个题是很常见的题目，使用变量分别保存当前的字符以及其出现的次数，如果所有的字符都用完则没有下一个字符了。

注意两点：

1. 字符出现的次数可能>=10
2. 最后一个字符用完时才算结束

C++代码如下：

class StringIterator {
public:
    StringIterator(string compressedString) {
        str = compressedString;
        index = 0;
        cur = ' ';
        count = 0;
    }

    char next() {
        if (!hasNext()) {
            return ' ';
        }
        if (count != 0) {
            count --;
            return cur;
        }
        cur = str[index];
        index ++;
        while (str[index] >= '0' && str[index] <= '9') {
            count = 10 * count + str[index] - '0';
            index ++;
        }
        count --;
        return cur;
    }

    bool hasNext() {
        return index < str.size() || count != 0;
    }
private:
    string str;
    int index;
    char cur;
    int count;
};

/**
 * Your StringIterator object will be instantiated and called as such:
 * StringIterator* obj = new StringIterator(compressedString);
 * char param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */

日期

2019 年 9 月 19 日 —— 举杯邀明月，对影成三人