回家续上刷题，题目集LeetCode 热题100

160. 相交链表

最笨的办法，挨个查

var getIntersectionNode = function(headA, headB) {
    let temp = headA;
    while(temp !==null) {
        let temp2 = headB 
        let interval = null
        while(temp2!== null) {
            if(temp==temp2) {
                interval = temp2
                break;
            }
            temp2 = temp2.next
        }
        if(interval) {
            return interval
        }
        temp = temp.next
    }
    return null
};

利用哈希表，判断内部是否有重合

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getIntersectionNode = function(headA, headB) {
    let list = new Set();
    let temp1 = headA;
    while(temp1!=null) {
       list.add(temp1)
       temp1 = temp1.next
    }
    let temp2 = headB;
    while(temp2!=null){
        if(list.has(temp2)){
            return temp2
        }
        temp2 = temp2.next
    }
   return  null
};

学习到一个新的解决两个链表长度不一样的遍历方法

案例 headA=[1,9,1,2,4] headB=[3,2,4]==>headB只有3个元素，3个元素遍历完 headB=[1,9,1,2,4] headA=[2,4]==>headA遍历完headA=[3,2,4] headB=[1,2,4] 按正常条件判断第二个元素条件满足跳出循环

var getIntersectionNode = function(headA, headB) {
    if (headA === null || headB === null) {
        return null;
    }
    let pA = headA, pB = headB;
    while (pA !== pB) {
        pA = pA === null ? headB : pA.next;
        pB = pB === null ? headA : pB.next;
    }
    return pA;
};

滴滴面试题

上一年滴滴一面有一个手写题：‘判断链表是不是环’，这个题目跟160相似

206、反转链表

var reverseList = function(head) {
    let resover = null
    let cur=head
    while(cur){
        const next =cur.next
        cur.next = resover
        resover=cur
        cur=next
    }
    return resover
};

经典面试题 2、两数相加

涉及进位问题

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function(l1, l2) {
    let head = null, tail = null;
    let carry = 0;
    while (l1 || l2) {
        const n1 = l1 ? l1.val : 0;
        const n2 = l2 ? l2.val : 0;
        const sum = n1 + n2 + carry;
        if (!head) {
            head = tail = new ListNode(sum % 10);
        } else {
            tail.next = new ListNode(sum % 10);
            tail = tail.next;
        }
        carry = Math.floor(sum / 10);
        if (l1) {
            l1 = l1.next;
        }
        if (l2) {
            l2 = l2.next;
        }
    }
    if (carry > 0) {
        tail.next = new ListNode(carry);
    }
    return head;
};

146LRU

这个也是一个遇见过一次面试出的手写题，力扣的这个算一个简易版的LRU缓存实现，以前就应该多刷刷力扣，呜呜🥹