面试11之以给定值x为基准将链表分割成两部分

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编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前

给定一个链表的头指针 ListNode* pHead，请返回重新排列后的链表的头指针。注意：分割以后保持原来的数据顺序不变。

分析：（1）创建两个链表，把大于等于x的放到链表big,小于x的放到链表small

（2）将两个链表拼接起来,注意处理链表的头尾结点。

#include<iostream>using namespace std;#include<vector>struct ListNode{int val;struct ListNode *next;ListNode(int x):val(x),next(NULL) { }};ListNode* Partition(ListNode*pHead,int x){if(pHead == NULL)return NULL;ListNode *smallstart = NULL;ListNode *smallend = NULL;ListNode *bigstart = NULL;ListNode *bigend = NULL;ListNode *pCur = pHead;while(pCur){if (pCur->val < x){if (smallend == NULL) //表示这是小的部分的头结点{smallstart = smallend = pCur;}else{smallend->next = pCur;smallend = pCur;}}else{if(bigend == NULL) //表示这是大于或等于x的头结点{bigstart = bigend = pCur;}else{bigend->next = pCur;bigend = pCur;}}pCur = pCur->next;}if(smallend != NULL){smallend->next = bigstart;}if(bigend != NULL){bigend->next = NULL;}return smallstart == NULL ? bigstart: smallstart;}void test(){ListNode *p1 = new ListNode(1);ListNode *p2 = new ListNode(8);ListNode *p3 = new ListNode(3);ListNode *p4 = new ListNode(2);ListNode *p5 = new ListNode(7);p1->next = p2;p2->next = p3;p3->next = p4;p4->next = p5;ListNode *pHead =Partition(p1,3);while (pHead){cout << pHead->val << " ";pHead = pHead->next;}cout <<endl;delete p1;delete p2;delete p3;delete p4;delete p5;}int main(){test();cout << "hello..." <<endl;return 0;}

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