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Intersection of Two Linked Lists
该题目对内存的使用极其变态。所用变量不能超过4个。否侧会内存超限。解法有两个，其中第二种解法，内存还需要优化，否则会内存越界。
解法一：
class Solution {public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode* tempA = headA;ListNode* tempB = headB; int countA=0,countB=0; if(headA == NULL || headB == NULL) return NULL; while(tempA->next != NULL) { countA++; tempA = tempA->next; } while(tempB->next != NULL) { countB++; tempB = tempB->next; } if(tempA != tempB) return NULL; if(countA > countB) { int count = countA- countB; while(count>0) { headA = headA->next; count--; } } else { int count = countB - countA; while(count>0) { headB = headB->next; count--; } } while(headA != headB) { headA = headA->next; headB = headB->next; } return headA; } };
解法二：
class Solution {public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode * endA; if(headA == NULL || headB == NULL) return NULL; endA = getEnd(headA); endA->next = headA; bool ret ; ListNode * pmeetpoint = NULL; ret = RoundCircle(headB,&pmeetpoint); if(ret == false) return NULL; ListNode *presult; presult = GetEntry(headB,pmeetpoint); endA->next = NULL; return presult; } ListNode* GetEntry(ListNode *headA, ListNode *pmeetpoint) { while(headA != pmeetpoint) { headA = headA->next; pmeetpoint = pmeetpoint->next; } return headA; } ListNode * getEnd(ListNode *headA) { while(headA->next != NULL) { headA= headA->next; } return headA; } bool RoundCircle(ListNode *headB,ListNode** ppmeetpoint) { ListNode * pslow,*pquick; pslow = headB; pquick = headB; while(1) { if(pquick->next ==NULL) return false; pquick = pquick->next; if(pquick->next ==NULL) return false; pquick = pquick->next; pslow = pslow->next; if(pslow == pquick) { *ppmeetpoint = pslow; return true; } } }};
Intersection of Two Linked Lists
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