Intersection of Two Linked Lists | 28.9% | |||
Sort List | 2013-11-16 | 20.8% | Medium | |
Insertion Sort List | 2013-11-12 | 25.4% | Medium | |
Reorder List | 2013-11-02 | 20.5% | Medium | |
Linked List Cycle II | 2013-10-30 | 30.8% | Medium | |
Linked List Cycle | 2013-10-28 | 35.9% | Medium | |
Copy List with Random Pointer | 2013-10-03 | 23.6% | Hard | |
Flatten Binary Tree to Linked List | 2012-10-14 | 28.2% | Medium | |
Convert Sorted List to Binary Search Tree | 2012-10-02 | 27.4% | Medium | |
Reverse Linked List II | 2012-06-27 | 26.1% | Medium | |
Partition List | 2012-04-30 | 27.0% | Medium | |
Remove Duplicates from Sorted List II | 2012-04-22 | 24.8% | Medium | |
Remove Duplicates from Sorted List | 2012-04-22 | 34.6% | Easy | |
Merge Two Sorted Lists | 2012-03-30 | 33.1% | Easy | |
Rotate List | 2012-03-27 | 21.9% | Medium | |
Merge k Sorted Lists | 2012-02-13 | 21.0% | Hard | |
Remove Nth Node From End of List | 2012-01-27 | 28.9% | Easy |
class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode * p=headA, *pp=NULL, *q=headB, *pq=NULL, *tmp; int i = 0; while(p) i++, pp = p, p=p->next; int j = 0; while(q) j++, pq = q, q=q->next; if(pp!=pq) return NULL; if(i<j) tmp = headA, headA = headB, headB = tmp; for(int k = 0; k<abs(i-j); k++) headA=headA->next; while(headA!=headB) headA=headA->next, headB=headB->next; return headA; } };
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *sortList(ListNode *head) { if(!head) return head; return sort(head, NULL); } ListNode* sort(ListNode *l, ListNode *r){ // assert(l != NULL); if( l ->next == r ) { l->next = NULL; return l; } ListNode *p=l, *p2=l; while(p2 != r && p2->next != r){ p=p->next; p2=p2->next->next; } ListNode *a, *b, *head = NULL, *tail = NULL; a = sort(l, p); b = sort(p, r); while(a != NULL && b != NULL ){ if( a->val > b->val ) {ListNode *tmp = a; a = b; b = tmp; }//@error if(!head) head = a; if(tail) tail -> next = a; tail = a; a=a->next; } if( a!= NULL ){ if(!head) head = a; if(tail) tail -> next = a; } if( b!= NULL){ if(!head) head = b; if(tail) tail -> next = b; } return head; } };
上述代码要注意
while(p2!=r && p2->next!=r){ p=p->next; p2=p2->next->next; }
这里p = mid 指向了中点(l+r)/2,注意r是不包括在内的,而 p2 指向了左闭右开 【l, r) 区间的最后一个元素,或者r。注意如果把【l,r)改写成全闭区间【L,R], 那么上面的运算的结果mid=( L + R+1)/2。总之mid实际上是中间偏右的一个点,不是纯正的中点。
在此基础上,就可以二分成: sort(l, r) => sort(l, p), sort(p, r) ([l, r) => [l, (l+r)/2), [(l+r)/2, r) )
上述代码还可以优化: 对于当前链表<head, tail>,向后链入元素a,只需要以下三句:
if(!head) head = a; else tail -> next = a; tail = a;
是不是逻辑很清晰?
我们就把上面三句话构成的append方法叫做 head-tail-append吧。
上题也可以基数排序,毕竟保存的都是数嘛。注意弄19个链表<lists[i], tails[i]>;
从个位往高位依次排序,重复下面过程:
下面是非常漂亮的一段实现代码。
ListNode *sortList(ListNode *head) { ListNode* lsts[19];//-9~9 ListNode* tails[19]; typedef ListNode* PNode; ListNode* lst=head,*tail; if(lst==NULL) return NULL; //基数排序 int d=1; while(d<0xfffffff){ for(int i=0;i<19;i++) lsts[i]=NULL; for(PNode it=lst;it!=NULL;it=it->next){ int s=9+(it->val/d)%10;//0~18 // head-tail-append 三句代码,append 节点it if(!lsts[s]) lsts[s]=it; else tails[s]->next=it; tails[s]=it; } lst=NULL; // <lst, tail> inited as <NULL, xx> for(int i=0;i<19;i++){ if(lsts[i]){ // head-tail-append 三句代码再次出现 if(!lst) lst=lsts[i],tail=tails[i]; else tail->next=lsts[i],tail=tails[i]; } } tail->next=NULL; d*=10; } return lst; }
步骤和上面类似:
1 二分 dfs(l, r):
寻找mid的方法(下面方法找到的mid偏右)
p1 = l, p2 = l; while(p2 !=r && p2->next !=r){ p1 = p1->next, p2 = p2->next->next; } //p1 将指向[l, r) 中的(l+r)/2 //一定要记得二分法中mid=(l+r)/2的含义: // 示意图:# # * # #, 或者 # * # // 区间对应[l, r] 或 [l, r)
TreeNode *convert(ListNode *l, ListNode *r){ if(l == r) return NULL; if(r == l->next) return new TreeNode(l->val); ListNode *mid=l, *p2 = l; while(p2 != r && p2->next != r){ mid = mid->next, p2 = p2->next->next; } TreeNode *lt = convert(l, mid), *rt = convert(mid->next, r); TreeNode *p = new TreeNode(mid->val); p->left = lt, p->right = rt; return p; }
或者用计数法找中点mid,则n=区间节点个数,区间[0, n-1]的中点mid的下标为( 0 + n-1 ) / 2, 用 pm = left; for(i=0; i<mid; i++) pm = pm->next;即可得到中央节点pm
TreeNode *dfs(ListNode *left,int n){ if(n<=0) return NULL; ListNode *mid=left; for(int i=0;i<(n-1)/2;i++) mid=mid->next; TreeNode *node=new TreeNode(mid->val); TreeNode *pl=dfs(left,(n-1)/2),*pr=dfs(mid->next,n/2); node->left=pl,node->right=pr; return node; }
class Solution { public: ListNode *removeNthFromEnd(ListNode *head, int n) { ListNode *p=head,*pre = NULL, *q = head; int i = 1; for(; i<n && q; i++) q = q->next; if(i==n){ while(q->next) pre = p, p = p->next, q = q->next; if(!pre){ head = head -> next; } else pre->next = pre->next->next; free(p); } return head; } };或者
class Solution { public: ListNode *removeNthFromEnd(ListNode *head, int n) { ListNode *p=head,*pre; int i;for(i=0;p!=NULL;i++){ p=p->next;} //i=size if(n>i||n<=0) return head; n=i-n; //point[n] pre=head; for(i=0;i<n-1;i++) pre=pre->next; if(n-1<0) {head=head->next; return head;} pre->next=pre->next->next; return head; } };
class Solution { public: ListNode *partition(ListNode *head, int x) { ListNode *p=head,*pre=NULL; while(p!=NULL&&p->val<x){ pre=p,p=p->next;} if(p!=NULL){//p->val>=x ListNode *q=p->next,*preq=p; while(q!=NULL){ if(q->val<x){ //cut q and insert before p preq->next=q->next; if(pre!=NULL) {pre->next=q;} else {head=q;}//head change q->next=p; pre=q; } preq=q,q=q->next;//@attention: not good } } return head; } };或者
class Solution { public: ListNode *partition(ListNode *head, int x) { ListNode * p = head, *pp = NULL; while(p && p->val < x) pp = p, p = p->next; if(p){ ListNode * q = p->next, * pq = p; while(q){ if( q->val < x ) { pq -> next = pq->next->next; if(!pp) head = q; else pp -> next = q; pp = q; pp -> next = p; q = pq -> next;//@important } else pq = q, q = q->next; } } return head; } };
整体上分为三步骤:
第一步对每个节点拷贝一个新的跟在后面,
第二步非常重要:在这里而不是在后面修改新节点的random指针,并且注意“空指针”
第三步,拆分原有链表和新增链表(用p1,q1分别表示新增表头尾指针即可)。用到了上述的head-tail-append三句代码。
/** * Definition for singly-linked list with a random pointer. * struct RandomListNode { * int label; * RandomListNode *next, *random; * RandomListNode(int x) : label(x), next(NULL), random(NULL) {} * }; */ class Solution { public: RandomListNode *copyRandomList(RandomListNode *head) { RandomListNode * p = head, *np; while(p){ RandomListNode * t = new RandomListNode(p->label); t->random = p->random; t->next = p->next; p->next = t; p = t->next; } //@error: should change random below, not later p = head; while(p){ np = p->next; if(np->random) np->random = np->random->next; p = np->next; } p = head; RandomListNode *p1=NULL, *q1; while(p){ np = p->next; // if(np->random) np -> random = np->random->next; //@error: should be changed earlier if(p1) q1->next = np; else p1 = np; q1 = np; p -> next = np->next; p = p -> next; } return p1; } };
class Solution { public: ListNode *insertionSortList(ListNode *head) { if(!head) return head; ListNode *q =NULL, *pre = NULL, *p = head->next, *np; head -> next = NULL; while(p){ np = p->next; p->next = NULL; // cut p pre = NULL, q = head; while(q && q->val <= p->val) pre = q, q = q->next; if(!pre) p->next = head, head = p; else p -> next = pre -> next, pre->next = p; p = np; } return head; } };
只想到了O(n)空间复杂度的解法。有O(1)空间的解法吗?
void reorderList(ListNode *head) { if(!head) return; vector<ListNode *> vec; ListNode * p =head; while(p) vec.push_back(p), p = p->next; int n = vec.size(); for(int i = 0; i<=(n-1)/2; i++){ if(i>0) vec[n-i]->next = vec[i]; vec[i]->next = vec[n-1-i]; vec[n-1-i]->next = NULL; } }
Remove Duplicate 2
class Solution { public: ListNode *deleteDuplicates(ListNode *head) { ListNode * pre = NULL, * p = head, *np; head = NULL; // <head, pre> inited to <NULL, xx> while(p){ ListNode * q = p->next, *tmp; while(q && q->val == p->val) q = q->next; // append p to <head, pre> if(q==p->next){ p->next = NULL; if(!pre) head = p; else pre->next = p; pre = p; } //skip duplicated p else{ while(p!=q) { ListNode *tmp = p->next; free(p); p = tmp; } } //next p p = q; } return head; } };
保持左边<head, pre>和右边的p的链接。
ListNode *deleteDuplicates(ListNode *head) { ListNode *p=head,*pre=NULL; while(p!=NULL&&p->next!=NULL){ if(p->val==p->next->val){ int val=p->val; ListNode *next; while(p!=NULL&&p->val==val){ next=p->next; delete p; p=next; } // <head, pre> link to next p if(pre!=NULL) pre->next=p; else { head=p; } } else{ // p append to <head, pre> pre=p,p=p->next; } } return head; } };
注意np,pp分别代表什么
ListNode *swapPairs(ListNode *head) { ListNode * p = head, *q = p, *np, *pp = NULL; while(p && p->next){ np = p->next->next; q = p->next; q -> next = p; // link last section with q if(pp) pp -> next = q; // first section, then head change else head = q; pp = p, p = np; } // link last section with p=NULL or single element if(pp) pp->next = p; return head; }
这个题目比较坑爹的地方在于k的大小可以超过n,
记得shift k相当于把下标为(n-k%n)%n的元素给作为首节点,就OK了
ListNode *rotateRight(ListNode *head, int k) { int n =0; ListNode * p = head, * pre = NULL, *tail = NULL; while(p){ n++; tail = p, p = p->next; } if(n==0) return head; k = (n - k%n)%n; if(k==0) return head; p = head; for(int i=0; i<k; i++){ pre = p, p = p->next; } tail -> next = head, head = p, pre -> next = NULL; return head; }
非常优美的,从第一个节点汪最后一个节点计算和append:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { // @error: no need to reverse l1 or l2 /*ListNode * p = l2, *pp =NULL; while(p){ ListNode *np = p->next; p->next = pp; pp = p, p = np; } l2 = pp;*/ ListNode * l3 = NULL, *tail;//head int c =0; while(l1 || l2 || c){ int a= l1? l1->val:0; int b = l2? l2->val:0; if(l1) l1 = l1->next; if(l2) l2 = l2->next; int t = a + b + c; if(t>=10) t%=10, c=1; else c = 0; ListNode * tmp = new ListNode(t); //tmp->next = l3, l3 = tmp; //@error: should not insert in the head of <l3, xx>, because head is the lowest bit if(!l3) l3 = tmp; //@attention: should append tmp to <l3, tail> else tail->next = tmp; tail = tmp; } return l3; }