问题
思路
cur = pre->next; post = cur=>next; cur->next = pre; cur = post;
代码
ListNode* reverseList(ListNode *root) { if (root == NULL) return root; ListNode *cur = root; ListNode *pre = NULL; ListNode *post = NULL; ListNode *revRoot = NULL; while (cur != NULL) { post = cur->next; if (post == NULL) revRoot = cur; cur->next = pre; pre = cur; cur = post; } return revRoot; }
完整执行
View Code
结果
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本文转自jihite博客园博客，原文链接：http://www.cnblogs.com/kaituorensheng/p/3524888.html，如需转载请自行联系原作者
