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604. 迭代压缩字符串

English Version

题目描述

设计并实现一个迭代压缩字符串的数据结构。给定的压缩字符串的形式是，每个字母后面紧跟一个正整数，表示该字母在原始未压缩字符串中出现的次数。

设计一个数据结构，它支持如下两种操作： next 和 hasNext。

  • next() - 如果原始字符串中仍有未压缩字符，则返回下一个字符，否则返回空格。
  • hasNext() - 如果原始字符串中存在未压缩的的字母，则返回true，否则返回false。

 

示例 1：

输入：
["StringIterator", "next", "next", "next", "next", "next", "next", "hasNext", "next", "hasNext"]
[["L1e2t1C1o1d1e1"], [], [], [], [], [], [], [], [], []]
输出：
[null, "L", "e", "e", "t", "C", "o", true, "d", true]

解释：
StringIterator stringIterator = new StringIterator("L1e2t1C1o1d1e1");
stringIterator.next(); // 返回 "L"
stringIterator.next(); // 返回 "e"
stringIterator.next(); // 返回 "e"
stringIterator.next(); // 返回 "t"
stringIterator.next(); // 返回 "C"
stringIterator.next(); // 返回 "o"
stringIterator.hasNext(); // 返回 True
stringIterator.next(); // 返回 "d"
stringIterator.hasNext(); // 返回 True

 

提示：

  • 1 <= compressedString.length <= 1000
  • compressedString 由小写字母、大写字母和数字组成。
  • 在 compressedString 中，单个字符的重复次数在 [1,10 ^9] 范围内。
  • next 和 hasNext 的操作数最多为 100 。

解法

方法一：解析存储

将 compressedString 解析成字符 $c$ 和对应的重复次数 $x$，存储在数组或列表 $d$ 中，用 $p$ 指向当前字符。

然后在 next 和 hasNext 中进行操作。

初始化的时间复杂度为 $O(n)$，其余操作的时间复杂度为 $O(1)$。其中 $n$ 为 compressedString 的长度。

Python3

class StringIterator:

    def __init__(self, compressedString: str):
        self.d = []
        self.p = 0
        n = len(compressedString)
        i = 0
        while i < n:
            c = compressedString[i]
            x = 0
            i += 1
            while i < n and compressedString[i].isdigit():
                x = x * 10 + int(compressedString[i])
                i += 1
            self.d.append([c, x])

    def next(self) -> str:
        if not self.hasNext():
            return ' '
        ans = self.d[self.p][0]
        self.d[self.p][1] -= 1
        if self.d[self.p][1] == 0:
            self.p += 1
        return ans

    def hasNext(self) -> bool:
        return self.p < len(self.d) and self.d[self.p][1] > 0


# Your StringIterator object will be instantiated and called as such:
# obj = StringIterator(compressedString)
# param_1 = obj.next()
# param_2 = obj.hasNext()

Java

class StringIterator {
    private List<Node> d = new ArrayList<>();
    private int p;

    public StringIterator(String compressedString) {
        int n = compressedString.length();
        int i = 0;
        while (i < n) {
            char c = compressedString.charAt(i);
            int x = 0;
            while (++i < n && Character.isDigit(compressedString.charAt(i))) {
                x = x * 10 + (compressedString.charAt(i) - '0');
            }
            d.add(new Node(c, x));
        }
    }

    public char next() {
        if (!hasNext()) {
            return ' ';
        }
        char ans = d.get(p).c;
        if (--d.get(p).x == 0) {
            ++p;
        }
        return ans;
    }

    public boolean hasNext() {
        return p < d.size() && d.get(p).x > 0;
    }
}

class Node {
    char c;
    int x;

    Node(char c, int x) {
        this.c = c;
        this.x = x;
    }
}

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

C++

class StringIterator {
public:
    StringIterator(string compressedString) {
        int n = compressedString.size();
        int i = 0;
        while (i < n) {
            char c = compressedString[i];
            int x = 0;
            while (++i < n && isdigit(compressedString[i])) {
                x = x * 10 + (compressedString[i] - '0');
            }
            d.push_back({c, x});
        }
    }

    char next() {
        if (!hasNext()) return ' ';
        char ans = d[p].first;
        if (--d[p].second == 0) {
            ++p;
        }
        return ans;
    }

    bool hasNext() {
        return p < d.size() && d[p].second > 0;
    }
private:
    vector<pair<char, int>> d;
    int p = 0;
};

/**
 * 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();
 */

Go

type pair struct {
	c byte
	x int
}

type StringIterator struct {
	d []pair
	p int
}

func Constructor(compressedString string) StringIterator {
	n := len(compressedString)
	i := 0
	d := []pair{}
	for i < n {
		c := compressedString[i]
		x := 0
		i++
		for i < n && compressedString[i] >= '0' && compressedString[i] <= '9' {
			x = x*10 + int(compressedString[i]-'0')
			i++
		}
		d = append(d, pair{c, x})
	}
	return StringIterator{d, 0}
}

func (this *StringIterator) Next() byte {
	if !this.HasNext() {
		return ' '
	}
	ans := this.d[this.p].c
	this.d[this.p].x--
	if this.d[this.p].x == 0 {
		this.p++
	}
	return ans
}

func (this *StringIterator) HasNext() bool {
	return this.p < len(this.d) && this.d[this.p].x > 0
}

/**
 * Your StringIterator object will be instantiated and called as such:
 * obj := Constructor(compressedString);
 * param_1 := obj.Next();
 * param_2 := obj.HasNext();
 */

...