迭代器失效问题(转)
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近期看书遇到了迭代器失效的问题,认真搜了一些资料,也看了一些博客,感觉别人已经说得很详细了,就不重复造轮子了,贴上他人博客内容,留在参考学习。
原文地址:http://www.cppblog.com/JohnsonLu/archive/2009/10/14/98590.html
从一个简单的问题开始,删除数组中某个元素后连续重复的元素,例如 1,1,2,3,3,1,1,1,4,0 ---> 1, 2,3,1,4,0。
考虑了几秒,然后就开始动手写代码了:
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可是编译一下,出来一大堆error,仔细看一下出错信息,哦,原来自己忘记了,erase容器中元素的时候,迭代器会失效。。。顿时一身冷汗,自己平时迭代容器的时候,一般都保存了容器的end元素,要是此时迭代器失效。。。
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于是找到收藏的Effective STL,翻开条款9,找到了erase容器中元素的原则。以前曾经看过,不过年深日久,早就忘得一干二净了。现在还是把要点总结一下,记在blog上,供以后参考。
1. 对于关联容器(如map, set, multimap,multiset),删除当前的iterator,仅仅会使当前的iterator失效,只要在erase时,递增当前iterator即可。这是因为map之类的容器,使用了红黑树来实现,插入、删除一个结点不会对其他结点造成影响。
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因为iter传给erase方法的是一个副本,iter++会指向下一个元素。
2. 对于序列式容器(如vector,deque),删除当前的iterator会使后面所有元素的iterator都失效。这是因为vetor,deque使用了连续分配的内存,删除一个元素导致后面所有的元素会向前移动一个位置。还好erase方法可以返回下一个有效的iterator。
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3. 对于list来说,它使用了不连续分配的内存,并且它的erase方法也会返回下一个有效的iterator,因此上面两种方法都可以使用。
最后给出开始那个问题的一个正确的实现:
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PS. 不过实际上这个问题,用vector来实现不是很适合,因为每次删除一个元素,都会引起vector的一个resize操作。resize的时间复杂度是O(n),整个的resize操作要花费O(n^2)。最好是选择list最为容器,list最适合那些需要在容器中间做插入、删除的例子。
PS2. 另外一个可能的方法是remove_if + erase惯用法。remove_if的时间复杂度是O(n). erase的时间复杂度也是O(n),所以整个操作可以在O(n)时间里完成。(当然,对于这个例子,ShouldDeletePred比较难写)
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