604. Design Compressed String Iterator

Description

Design and implement a data structure for a compressed string iterator. 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.

Implement the StringIterator class:

  • next() Returns the next character if the original string still has uncompressed characters, otherwise returns a white space.
  • hasNext() Returns true if there is any letter needs to be uncompressed in the original string, otherwise returns false.

 

Example 1:

Input
["StringIterator", "next", "next", "next", "next", "next", "next", "hasNext", "next", "hasNext"]
[["L1e2t1C1o1d1e1"], [], [], [], [], [], [], [], [], []]
Output
[null, "L", "e", "e", "t", "C", "o", true, "d", true]

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

 

Constraints:

  • 1 <= compressedString.length <= 1000
  • compressedString consists of lower-case an upper-case English letters and digits.
  • The number of a single character repetitions in compressedString is in the range [1, 10^9]
  • At most 100 calls will be made to next and hasNext.

Solutions

Solution 1

Python Code
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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 Code
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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++ Code
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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 Code
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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();
 */