求链表相交时的交点（C语言）

//链表相交时的交点 PSListNode IntersectionNode(PSListNode pL1, PSListNode pL2) {int count1 = 0;int count2 = 0;PSListNode PSList1 = pL1;PSListNode PSList2 = pL2;PSListNode pMeetNode1 = HasCycle(pL1);PSListNode pMeetNode2 = HasCycle(pL2);if ((NULL == pL1) || (NULL == pL2)){return NULL;}else{//先求每个链表的长度//两个链表都没环if ((NULL == pMeetNode1) && (NULL == pMeetNode2)){while (NULL != PSList1){PSList1 = PSList1->pNextNode;count1++;}while (NULL != PSList2){PSList2 = PSList2->pNextNode;count2++;}}//两个链表都有环else if ((NULL != pMeetNode1) && (NULL != pMeetNode2)){PSListNode pInNode1 = FindEnterNode(PSList1, pMeetNode1);PSListNode pInNode2 = FindEnterNode(PSList2, pMeetNode2);//先计算头指针到环入口结点的长度，再计算环的长度while (PSList1 != pInNode1){PSList1 = PSList1->pNextNode;count1++;}while (PSList1->pNextNode != PSList1){PSList1 = PSList1->pNextNode;count1++;}count1++;;while (PSList2 != pInNode2){PSList2 = PSList2->pNextNode;count2++;}while (PSList2->pNextNode != PSList1){PSList2 = PSList2->pNextNode;count2++;}count2++;;}//一个有环，一个没环，不会相交else{return NULL;}//让长度长的链表的头指针先走它长于另一个链表的结点数//在计算链表长度时修改了这两个指针的值，在这儿需要把它们改回来PSList1 = pL1;PSList2 = pL2;if (count1 > count2){int temp = count1 - count2;while (0 == temp--){PSList1 = PSList1->pNextNode;}}else{int temp = count2 - count1;while (0 == temp--){PSList2 = PSList2->pNextNode;}}//此时，让两个链表的头指针同时移动，直到它们相等就找到了交点//因为题目是找交点，那么交点就存在，所以这儿不用怕死循环while (1){if (PSList1 = PSList2){break;}PSList1 = PSList1->pNextNode;PSList2 = PSList2->pNextNode;}return PSList1;} }

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