小心刪除容器中元素時的迭代器失效
阿新 • • 發佈:2019-01-10
從一個簡單的問題開始,刪除陣列中某個元素後連續重複的元素,例如 1,1,2,3,3,1,1,1,4,0 ---> 1, 2,3,1,4,0。
考慮了幾秒,然後就開始動手寫程式碼了:
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可是編譯一下,出來一大堆error,仔細看一下出錯資訊,哦,原來自己忘記了,erase容器中元素的時候,迭代器會失效。。。頓時一身冷汗,自己平時迭代容器的時候,一般都儲存了容器的end元素,要是此時迭代器失效。。。
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於是找到收藏的Effective STL,翻開條款9,找到了erase容器中元素的原則。以前曾經看過,不過年深日久,早就忘得一乾二淨了。現在還是把要點總結一下,記在blog上,供以後參考。
1. 對於關聯容器(如map, set, multimap,multiset),刪除當前的iterator,僅僅會使當前的iterator失效,只要在erase時,遞增當前iterator即可。這是因為map之類的容器,使用了紅黑樹來實現,插入、刪除一個結點不會對其他結點造成影響。
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因為iter傳給erase方法的是一個副本,iter++會指向下一個元素。
2. 對於序列式容器(如vector,deque),刪除當前的iterator會使後面所有元素的iterator都失效。這是因為vetor,deque使用了連續分配的記憶體,刪除一個元素導致後面所有的元素會向前移動一個位置。還好erase方法可以返回下一個有效的iterator。
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3. 對於list來說,它使用了不連續分配的記憶體,並且它的erase方法也會返回下一個有效的iterator,因此上面兩種方法都可以使用。
最後給出開始那個問題的一個正確的實現:
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PS. 不過實際上這個問題,用vector來實現不是很適合,因為每次刪除一個元素,都會引起vector的一個resize操作。resize的時間複雜度是O(n),整個的resize操作要花費O(n^2)。最好是選擇list最為容器,list最適合那些需要在容器中間做插入、刪除的例子。
PS2. 另外一個可能的方法是remove_if + erase慣用法。remove_if的時間複雜度是O(n). erase的時間複雜度也是O(n),所以整個操作可以在O(n)時間裡完成。(當然,對於這個例子,ShouldDeletePred比較難寫)
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