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                # 一、概念 1.對象的多重數組表示:對一組具有相同域的對象,每一個對象都可以用一個數組來表示 ![](https://box.kancloud.cn/2016-02-02_56b02bd08e4b5.gif) 2.對象的單數組表示:一個對象占據存儲中的一組連續位置 ![](https://box.kancloud.cn/2016-02-02_56b02bd09dadc.gif) # 二、代碼 ~~~ int Allocate_Object() { if(free == NULL) { cout<<"error:out of space"<<endl; exit(0); } else { int x = free; free = next[x]; return x; } } void Free_Object(int x) { next[x] = free; free = x; } ~~~ # 三、練習 10.3-1 多重數組 key :13, 4, 8, 19, 5, 11 next:1, 2, 3, 4, 5, -1 pre:-1, 0, 1, 2, 3, 4 單數組: 13, 3, -1,?4, 6, 3, 8, 9, 6, 19, 12, 9, 5, 15, 12, 11, 18, 15 10.3-2 ~~~ ALLOCATE-OBJECT() 1 if free = NIL 2 then error "out of space" 3 else x <- free 4 free <- A[x+1] 5 return x FREE-OBJECT(x) 1 A[x+1] <- free 2 free <- x ~~~ 10.3-3 在這里prev域沒有 在這里在這里prev域沒有意義,用不到 10.3-4 見[算法導論-10.3-4](http://blog.csdn.net/mishifangxiangdefeng/article/details/7707149) 假設當前的數組是緊湊的,即數組中有f個元素,都位于數組的前f個位置 分配一個新的元素時,把f+1的位置分配給它 刪除一個元素時,假設待刪除的元素的位置是i,先修改元素prev[i]的next指針和元素next[i]的prev指針,刪除這個元素。這里數組中間就留下一個空位,讓第f個元素填充這個空位,具體方法是修改元素prev[f]的next指針和元素next[f]和prev指針 10.3-5 與10.3-4類似
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