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Implement an iterator to flatten a 2d vector.
For example, Given 2d vector =
[
[1,2],
[3],
[4,5,6]
] By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,2,3,4,5,6].
分析：https://segmentfault.com/a/1190000003791233
用一个数组表示每个List的迭代器，然后再记录一个变量，用来表示当前用到了第几个迭代器。
1 public class Vector2D { 2 3 List<Iterator<Integer>> its; 4 int curr = 0; 5 6 public Vector2D(List<List<Integer>> vec2d) { 7 this.its = new ArrayList<Iterator<Integer>>(); 8 for (List<Integer> l : vec2d) { 9 // 只将非空的迭代器加入数组 10 if (l.size() > 0) { 11 this.its.add(l.iterator()); 12 } 13 } 14 } 15 16 public int next() { 17 Integer res = its.get(curr).next(); 18 // 如果该迭代器用完了，换到下一个 19 if (!its.get(curr).hasNext()) { 20 curr++; 21 } 22 return res; 23 } 24 25 public boolean hasNext() { 26 return curr < its.size() && its.get(curr).hasNext(); 27 } 28 }
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5739787.html
