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原题是这样的:
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
For example,
Given 1->4->3->2->5->2
and x = 3,
return 1->2->2->4->3->5
.
要求我们将比给定值小的值全部排在比给定值大或等于给定值的前面。
struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode* partition(ListNode* head, int x) { ListNode left(0), right(0); ListNode* l = &left; ListNdoe* r = &right; while (head) { ListNode* ret = (head->val < x) ? l : r; ret->next = head; ret = ret->next; // 尾部插入 head = head->next; } l->next = right.next; r.next = NULL; return left.next; } };