###問題描述
實現一個跳過null值的迭代器
[null,"1","2","3",null,"4",null]->["1","2","3","4"]
[null]->[]
[]->[]
###原來ide
Iterator<String> iterator = new Iterator<String>() { private Iterator<String> sourceIterator = source.iterator(); private String next; public boolean hasNext() { while (sourceIterator.hasNext()) { next = sourceIterator.next(); if (next != null) { return true; } } next = null; return false; } public String next() { while (next == null) { next = sourceIterator.next(); } String temp = next; next = null; return temp; } public void remove() { } };
我須要寫的如此複雜,才能保持Iterator特性,針對第一個case連續調用5次next方法拋出異常
###使用Guavacode
Iterator<String> iterator = new AbstractIterator<String>() { private Iterator<String> sourceIterator = source.iterator(); @Override protected String computeNext() { while (sourceIterator.hasNext()) { String next = sourceIterator.next(); if (next != null) { return next; } } return endOfData(); } };
只須要關注如何獲取下一個值就能夠了,不須要關注如何實現hasNext方法。
須要注意:AbstractIterator不支持remove方法rem