LeetCode 相交链表&环形链表II
题目链接：https://leetcode-cn.com/problems/intersection-of-two-linked-lists/
题目连接：https://leetcode-cn.com/problems/linked-list-cycle-ii/
题目大意
具体按左神书上的描述来实现，空间复杂度 O(1) 解决有环单链表相交问题。
分析
关于快慢指针能找到入环节点的一些说明强烈推荐这篇博客：https://blog.csdn.net/ffj0721/article/details/83302412
代码如下
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { 12 ListNode *loopA = detectCycle(headA), *loopB = detectCycle(headB); 13 int lenA, lenB; 14 15 if(loopA == NULL && loopB == NULL) { // 两个链表都没有环 16 lenA = getLen(headA); 17 lenB = getLen(headB); 18 19 // 长链表先走，然后再一起走 20 ListNode *pA = headA, *pB = headB; 21 if(lenA > lenB) for(int i = 0; i < lenA - lenB; ++i) pA = pA->next; 22 else for(int i = 0; i < lenB - lenA; ++i) pB = pB->next; 23 24 while(pA != pB) { 25 pA = pA->next; 26 pB = pB->next; 27 } 28 29 return pA; 30 } 31 else if(loopA != NULL && loopB != NULL) { // 两个链表都有环 32 if(loopA == loopB) { // 和都没有环的情况一样，只不过终点不是NULL了 33 lenA = getLen(headA, loopA); 34 lenB = getLen(headB, loopB); 35 36 ListNode *pA = headA, *pB = headB; 37 if(lenA > lenB) for(int i = 0; i < lenA - lenB; ++i) pA = pA->next; 38 else for(int i = 0; i < lenB - lenA; ++i) pB = pB->next; 39 40 while(pA != pB) { 41 pA = pA->next; 42 pB = pB->next; 43 } 44 45 return pA; 46 } 47 else { 48 ListNode *p = loopA->next; 49 50 while(p != loopA) { 51 if(p == loopB) return loopA; // 相交但入环点不同 52 p = p->next; 53 } 54 return NULL; // 不相交 55 } 56 } 57 return NULL; // 一个有环一个没环，一定不相交 58 } 59 60 // 如果链表无环，返回 NULL，否则返回第一个入环节点 61 ListNode *detectCycle(ListNode *head) { 62 if(head == NULL || head->next == NULL || head->next->next == NULL) return NULL; 63 64 ListNode *slow = head->next, *fast = head->next->next; 65 66 while(fast != slow) { 67 if(fast->next == NULL || fast->next->next == NULL) return NULL; 68 fast = fast->next->next; 69 slow = slow->next; 70 } 71 72 slow = head; 73 while(fast != slow) { 74 fast = fast->next; 75 slow = slow->next; 76 } 77 78 return slow; 79 } 80 81 int getLen(ListNode *head, ListNode *end = NULL) { 82 int ret = 0; 83 while(head != end) { 84 ++ret; 85 head = head->next; 86 } 87 return ret; 88 } 89 };
