類似本題的另外敘述題的解答
()[題目分析] 本題應先查找第i個結點記下第i個結點的指針然後從第i+個結點起直至第m(<i<m)個結點止依次插入到第i個結點之後最後將暫存的第i個結點的指針指向第m結點形成新的循環鏈表結束了倒置算法
LinkedList PatternInvert(LinkedList Lint im)∥L是有m個結點的鏈表的頭結點的指針表中從第i(<i<m)個結點到第m個結點構成循環部分鏈表本算法將這部分循環鏈表倒置
{if(i<|| i>=m || m<){printf(%d%d參數錯誤\nim);exit();}
p=L>next>next;∥p是工作指針初始指向第二結點(已假定i>)
pre=L>next;∥pre是前驅結點指針最終指向第i個結點
j=;∥計數器
while(j<i)∥查找第i個結點
{j++;pre=p;p=p>next;}∥查找結束p指向第i個結點
q=p;∥暫存第i個結點的指針
p=p>next;∥p指向第i+個結點准備逆置
j+=;∥上面while循環結束時j=i現從第i+結點開始逆置
while(j<=m)
{r=p>next;∥暫存p的後繼結點
p>next=pre>next;∥逆置p結點
pre>next=p;
p=r;∥p恢復為當前待逆置結點
j++;∥計數器增
}
q>next=pre>next;∥將原第i個結點的後繼指針指向原第m個結點
[算法討論] 算法中未深入討論imj的合法性因題目的條件是m>且<i<m因此控制循環並未用指針判斷(如一般情況下的p!=null)結束循環也未用指針判斷注意最後一句q>next=pre>next實現了從原第i個結點到原第m個結點的循環最後pre>next正是指向原第m個結點不可用p>next代替pre>next
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From:http://tw.wingwit.com/Article/program/sjjg/201311/23331.html
