前端手撕题总结篇（算法篇——来自Leetcode牛客）

链表

指定区域反转

找到区间（头和为 for循环当**时）->反转链表（返回反转过后的头和尾）->连接

function reverseBetween( head ,  m ,  n ) {//preEnd&cur&nextStart  cur.next断开if(m===n)return head;const vHeadNode=new ListNode(0);vHeadNode.next=head;let preEnd=null;let cur = vHeadNode;for(let i=0;i<n;i++){if(i===m-1)preEnd=cur;cur=cur.next;}let nextStart =cur.next;cur.next=null;const [start,end]=reverseList(preEnd.next);preEnd.next=start;end.next=nextStart;return vHeadNode.next;
}
function reverseList(head){const vHeadNode=new ListNode(-1);vHeadNode.next=head;let cur=head;while(cur){const next=cur.next;cur.next=vHeadNode.next;vHeadNode.next=cur;cur=next;}return [vHeadNode.next,head];
}

每K个一组反转（★★★）

指针preEnd&cur->while(cur)->curStart=cur;curEnd=preEnd,if(!preEnd)return vHead.next->反转连接-》cur=nextStart&preEnd=End(翻转过后）

function reverseKGroup(head, k) {// write code hereconst vHeadNode=new ListNode(-1);vHeadNode.next=head;let preEnd=vHeadNode;let cur=head;while(cur){let curStart=cur;let curEnd=preEnd;for(let i=0;i<k;i++){curEnd=curEnd.next;if(!curEnd)return vHeadNode.next;}let nextStart=curEnd.next;curEnd.next=null;const [start,end]=reverseSubList(curStart);preEnd.next=start;end.next=nextStart;cur=nextStart;preEnd=end;}return vHeadNode.next;
}
//反思一下自己吧！！！简单但请仔细
function reverseSubList(head) {let dummyNode = new ListNode(0);let cur = head;while (cur) {let next = cur.next;cur.next = dummyNode.next;dummyNode.next = cur;cur = next;}return [dummyNode.next, head];
}

合并K个已排序链表（★★★）

基础：合并两个已排序链表->归并排序

export function mergeKLists(lists: ListNode[]): ListNode {// // write code here// //nlogn --->归并// //拆开// //返回的是ListNode不是数组类型，ts编译会报错！太好了// if(lists.length<=1) return lists[0];// const mid=Math.floor(lists.length/2);// const left=mergeKLists(lists.slice(0,mid));// const right=mergeKLists(lists.slice(mid));// return merge(left,right);if(lists.length<=1)return lists[0];const mid =Math.floor(lists.length/2);const left = mergeKLists(lists.slice(0,mid));const right=mergeKLists(lists.slice(mid));return merge(left,right);
}function merge(pHead1:ListNode, pHead2:ListNode) {const dummyNode = new ListNode(0);let cur = dummyNode;while (pHead1 && pHead2) {if (pHead1.val < pHead2.val) {cur.next = pHead1;pHead1 = pHead1.next;} else {cur.next = pHead2;pHead2 = pHead2.next;}cur = cur.next;}cur.next=pHead1?pHead1:pHead2;return dummyNode.next;
}

基础

//双指针判断是否有环
function hasCycle( head ) {if(!head)return false;//起点相同let fast=head;let slow=head;//&&这里出错while(fast&&slow&&fast.next){//先跳后判断fast=fast.next.next;slow=slow.next;if(fast===slow)return true;}return false;}
//链表环的入口
function EntryNodeOfLoop(pHead)
{if(!pHead||!pHead.next)return null;let fast=pHead;let slow=pHead;while(fast&&slow&&fast.next){fast=fast.next.next;slow=slow.next;if(fast===slow){fast=pHead;//这！！！while(fast!==slow){fast=fast.next;slow=slow.next;}return fast;}}
}
//倒数第	K个节点
function FindKthToTail( pHead ,  k ) {// write code here//测试案例II,求长度！const genLen=(curNode)=>{let count=0;while(curNode){curNode=curNode.next;count++;}return count;}const len=genLen(pHead);//判断是否合理！！！if(k>len)return null;let fast=pHead,slow=pHead;while(k--){fast=fast.next;}//这条件！！！while(fast){fast=fast.next;slow=slow.next;}return slow;
}
//删除链表的倒数第n个节点(虚拟头结点）
function removeNthFromEnd( head ,  n ) {// write code herelet vHeadNode=new ListNode(0);vHeadNode.next=head;let fast=vHeadNode;let slow=vHeadNode;while(n--){fast=fast.next;}// //若不借助虚拟节点// //这一步真是太关键了// if(quick == null){//     return head.next// }//key:此处判断条件就不一样 fast.next&&slow.next！！！while(fast.next){fast=fast.next;slow=slow.next;}slow.next=slow.next.next;return vHeadNode.next;
}
//两链表第一个公共节点
export function FindFirstCommonNode(pHead1: ListNode, pHead2: ListNode): ListNode {//另一种思路：求长度差+先走+while相等退出返回//循环条件应该是当两个指针没有相遇时继续循环，即while(cur1 !== cur2)。//这样，当它们相等时（无论是相交节点还是都为null）就退出循环并返回cur1（此时等于cur2）if(!pHead1||!pHead2)return null;let cur1=pHead1;let cur2=pHead2;while(cur1!==cur2){cur1=cur1?cur1.next:pHead2;cur2=cur2?cur2.next:pHead1;}return cur2;
}

链表相加（★★★）

先反转+链表相加（大数相加）+反转结果

function addInList( head1 ,  head2 ) {// write code hereconst dummyNode=new ListNode(0);let cur=dummyNode;let reversedList1=reverseList(head1);let reversedList2=reverseList(head2);let carry=0;while(reversedList1||reversedList2||carry){let val1=reversedList1?reversedList1.val:0;let val2=reversedList2?reversedList2.val:0;const sum=val1+val2+carry;const digit=sum%10;const node =new ListNode(digit);carry=Math.floor(sum/10);cur.next=node;cur=cur.next;//!!!移动指针！！！链表相加肯定要移动指针!!!if(reversedList1)reversedList1=reversedList1.next;if(reversedList2)reversedList2=reversedList2.next;}return reverseList(dummyNode.next);
}
function reverseList(head){let vHeadNode=new ListNode(0);let cur=head;while(cur){let next=cur.next;cur.next=vHeadNode.next;vHeadNode.next=cur;cur=next;}return vHeadNode.next;
}

单链表排序

分治（归并）排序
key：找链表中点（需借助虚拟头结点）->返回条件->合并（两个已排序链表）

function sortInList( head ) {// write code here//分治if(!head||!head.next)return head;const middle=findMiddle(head);let right=middle.next;middle.next=null;const leftHead=sortInList(head);const rightHead=sortInList(right);const merge=(leftHead,rightHead)=>{const dummyNode=new ListNode(0);let cur=dummyNode;let head1=leftHead;let head2=rightHead;while(head1&&head2){if(head1.val>head2.val){cur.next=head2;head2=head2.next;}else{cur.next=head1;head1=head1.next;}cur=cur.next;}cur.next=head1?head1:head2;return dummyNode.next;}return merge(leftHead,rightHead);
}
function findMiddle(head){//要虚拟头结点!!!!这一个函数错了，导致整个通过不了！！！const dummyNode=new ListNode(0);dummyNode.next=head;let fast=dummyNode;let slow=dummyNode;while(fast&&fast.next){fast=fast.next.next;slow=slow.next;}return slow;
}

回文链表

找链表中节点（同上）-》反转左侧链表-》比较，若有不对等返回false

export function isPail(head: ListNode): boolean {// write code hereconst middle=findMiddle(head);let leftHead=head;let rightHead=middle.next;middle.next=null;const reverseList=(head:ListNode|null):ListNode=>{const dummyNode=new ListNode(0);let cur=head;while(cur){const next=cur.next;cur.next=dummyNode.next;dummyNode.next=cur;cur=next;}return dummyNode.next;}let reverseRightHead=reverseList(rightHead);while(leftHead&&reverseRightHead){if(leftHead.val!==reverseRightHead.val)return false;leftHead=leftHead.next;reverseRightHead=reverseRightHead.next;}return true;}
function findMiddle(head:ListNode):ListNode{const dummyNode=new ListNode(0);dummyNode.next=head;let fast=dummyNode;let slow=dummyNode;while(fast&&fast.next){fast=fast.next.next;slow=slow.next;}return slow;
}

链表奇偶排序

指针：odd、even&evenHead
条件：even&&even.head
返回：head

function oddEvenList( head ) {if(!head||!head.next)return head;// write code herelet odd=head;//奇数let even=head.next;//偶数let evenHead=even;//记录偶节点的头结点//终止条件：偶节点&偶节点.next不为空！！！！//even&&even.nextwhile(even&&even.next){odd.next=even.next;odd=odd.next;even.next=odd.next;even=even.next;}odd.next=evenHead;return head;
}

删除重复元素

I->cur=head

function deleteDuplicates( head ) {// write code herelet cur=head;//链表靠着时next指针，不要想着还计算长度什么！while(cur){while(cur.next&&cur.val===cur.next.val){cur.next=cur.next.next;}cur=cur.next;}return head;}

II(♥）

1. 虚拟头结点
2. 找到if(cur.next.val=cur.next.next.val)并存储至temp，内部while（cur.next&&cur.next.val=temp）删除节点
3. 否则就是cur=cur.next

export function deleteDuplicates(head: ListNode): ListNode {// // write code here// const vHeadNode=new ListNode(0);// vHeadNode.next=head;// let cur=vHeadNode;// let temp=0;// while(cur.next&&cur.next.next){//     if(cur.next.val===cur.next.next.val){//         temp=cur.next.val;//         //相同的节点都跳过//         while(cur.next&&cur.next.val===temp){//             cur.next=cur.next.next;//         }//     }else{//         //其他节点就向后移//         cur=cur.next;//     }// }// return vHeadNode.next;const vHeadNode=new ListNode(0);vHeadNode.next=head;//虚拟头结点，解决第一个和第二个元素就相同的情况let cur=vHeadNode;let temp=0;while(cur.next&&cur.next.next){if(cur.next.val===cur.next.next.val){temp=cur.next.val;while(cur.next&&cur.next.val===temp){cur.next=cur.next.next;}}else{cur=cur.next;}}return vHeadNode.next;
}