[LeetCode] 281. Zigzag Iterator 之字形迭代器

Given two 1d vectors, implement an iterator to return their elements alternately.

For example, given two 1d vectors:

v1 = [1, 2]
v2 = [3, 4, 5, 6]

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1, 3, 2, 4, 5, 6].

Follow up: What if you are given k 1d vectors? How well can your code be extended to such cases?

Clarification for the follow up question - Update (2015-09-18):
The "Zigzag" order is not clearly defined and is ambiguous for k > 2 cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example, given the following input:

[1,2,3]
[4,5,6,7]
[8,9]

It should return [1,4,8,2,5,9,3,6,7].

跟Flatten 2D Vector有些类似，那道题是横向打印，这道题是纵向打印，虽然方向不同，但是实现思路都是大同小异。

如果只有2个Vector，可以用两个指针来回切换两个数组。如果是k个，则需要用queue来存，然后再pop出来。

Java:

public class ZigzagIterator {
    List<Iterator<Integer> > iters = new ArrayList<Iterator<Integer> >(); 
    
    int count = 0;
 
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        if( !v1.isEmpty() ) iters.add(v1.iterator());
        if( !v2.isEmpty() ) iters.add(v2.iterator());
    }
 
    public int next() {
        int x = iters.get(count).next();
        if(!iters.get(count).hasNext()) iters.remove(count);
        else count++;
        
        if(iters.size()!=0) count %= iters.size();
        return x;
    }
 
    public boolean hasNext() {
        return !iters.isEmpty();
    }
}
 
/**
 * Your ZigzagIterator object will be instantiated and called as such:
 * ZigzagIterator i = new ZigzagIterator(v1, v2);
 * while (i.hasNext()) v[f()] = i.next();
 */　　

Java:

public class ZigzagIterator {

    public Queue<Iterator> queue;
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        queue = new LinkedList<>();
        if (v1.size() != 0) {
            queue.offer(v1.iterator());
        }
        if (v2.size() != 0) {
            queue.offer(v2.iterator());
        }
    }

    public int next() {
        hasNext();
        Iterator it = queue.poll();
        int val = (Integer)it.next();
        if (it.hasNext()) {
            queue.offer(it);
        }
        return val;
    }

    public boolean hasNext() {
        return !queue.isEmpty();
    }
}　　

Python:

# Time:  O(n)
# Space: O(k)
import collections

class ZigzagIterator(object):

    def __init__(self, v1, v2):
        """
        Initialize your q structure here.
        :type v1: List[int]
        :type v2: List[int]
        """
        self.q = collections.deque([(len(v), iter(v)) for v in (v1, v2) if v])

    def next(self):
        """
        :rtype: int
        """
        len, iter = self.q.popleft()
        if len > 1:
            self.q.append((len-1, iter))
        return next(iter)

    def hasNext(self):
        """
        :rtype: bool
        """
        return bool(self.q)

# Your ZigzagIterator object will be instantiated and called as such:
# i, v = ZigzagIterator(v1, v2), []
# while i.hasNext(): v.append(i.next())

C++:

class ZigzagIterator {
public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        v.push_back(v1);
        v.push_back(v2);
        i = j = 0;
    }
    int next() {
        return i <= j ? v[0][i++] : v[1][j++];
    }
    bool hasNext() {
        if (i >= v[0].size()) i = INT_MAX;
        if (j >= v[1].size()) j = INT_MAX;
        return i < v[0].size() || j < v[1].size();
    }
private:
    vector<vector<int>> v;
    int i, j;
};

C++:

class ZigzagIterator {
public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        int n1 = v1.size(), n2 = v2.size(), n = max(n1, n2);
        for (int i = 0; i < n; ++i) {
            if (i < n1) v.push_back(v1[i]);
            if (i < n2) v.push_back(v2[i]);
        }
    }
    int next() {
        return v[i++];
    }
    bool hasNext() {
        return i < v.size();
    }
private:
    vector<int> v;
    int i = 0;
};

C++: queue

class ZigzagIterator {
public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        if (!v1.empty()) q.push(make_pair(v1.begin(), v1.end()));
        if (!v2.empty()) q.push(make_pair(v2.begin(), v2.end()));
    }
    int next() {
        auto it = q.front().first, end = q.front().second;
        q.pop();
        if (it + 1 != end) q.push(make_pair(it + 1, end));
        return *it;
    }
    bool hasNext() {
        return !q.empty();
    }
private:
    queue<pair<vector<int>::iterator, vector<int>::iterator>> q;
};

　　

　　

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