24.两两交换链表中的节点
本题要点
要点看代码中的注释
代码实现
function swapPairs(head: ListNode | null): ListNode | null {
// 定义虚拟头节点
let dummyHead = new ListNode(0, head);
let cur = null, temp1 = null, temp2 = null, temp3 = null;
cur = dummyHead;
while(cur.next !== null && cur.next.next !== null){
temp1 = cur.next;
temp2 = cur.next.next;
temp3 = cur.next.next.next;
cur.next = cur.next.next;
temp2.next = temp1;
temp1.next = temp3;
// 指向下一个节点
cur = cur.next.next;
}
return dummyHead.next;
};
19.删除链表的倒数第N个节点
本题要点
cur表示较慢的指针,fast表示较快的指针。
fast前移n+1位后,cur和fast一起遍历,当fast指向null时,cur指向待移除元素的前一位,执行移除操作。
代码实现
var removeNthFromEnd = function (head, n) {
let dummyHead = new ListNode(0, head);
let cur = dummyHead
let fast = dummyHead
while(n-- && fast){
fast = fast.next;
}
//fast到末尾时(为null时),cur需要指向待删除元素前一位
fast = fast.next;
while(fast){
fast = fast.next;
cur = cur.next;
}
cur.next = cur.next.next;
return dummyHead.next;
};
面试题 02.07. 链表相交
本题要点
将两个链表尾端平齐之后,比较节点是否一样,节点一样的时候说明链表在此节点汇合成一个。
代码实现
var getIntersectionNode = function(headA, headB) {
let lenA = 0, lenB = 0;
let curA = headA, curB = headB;
//统计A B链表的长度
while(curA){
curA = curA.next;
lenA++;
}
while(curB){
curB = curB.next;
lenB++;
}
curA = headA;
curB = headB;
//解构赋值
if(lenB > lenA){
[lenA, lenB] = [lenB, lenA];
[curA, curB] = [curB, curA];
}
let gap = lenA - lenB;
while(gap--){
curA = curA.next;
}
while(curA && curB){
if(curA === curB){
return curA;
}
curA = curA.next;
curB = curB.next;
}
return null;
};
142.环形链表II
本题要点
代码实现
var detectCycle = function (head) {
let fast = head, slow = head;
while (fast && fast.next) {
fast = fast.next.next;
slow = slow.next;
if (fast === slow) {
//直接使用slow,让他们同时后移x(或者说z)步
slow = head;
while (fast !== slow) {
fast = fast.next;
slow = slow.next;
}
return fast;
}
}
return null;
};