[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:

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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:

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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:

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# 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++:

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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++:

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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

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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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