链表 - 链表相交

面试题 02.07. 链表相交

方法一：哈希集合

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

方法二：双指针

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
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;
};

方法三：链表对齐

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
var getListLen = function (head) {
    let len = 0, cur = head;
    while (cur) {
        len++;
        cur = cur.next;
    }
    return len;
};

var getIntersectionNode = function(headA, headB) {
    let curA = headA, curB = headB,
        lenA = getListLen(headA), // 求链表A的长度
        lenB = getListLen(headB);
    if (lenA < lenB) { // 让curA为最长链表的头，lenA为其长度
        // 交换变量注意加“分号”，两个数组交换变量在同一个作用域下时
        // 如果不加分号，下面两条代码等同于一条代码：[curA, curB] = [lenB, lenA]

        [curA, curB] = [curB, curA];
        [lenA, lenB] = [lenB, lenA];

    }
    let i = lenA - lenB; // 求长度差
    while (i-- > 0){  // 让curA和curB在同一起点上（末尾位置对齐）
        curA = curA.next;
    } 
    while (curA && curA !== curB) { // 变量curA和curB，遇到相同则直接返回
        curA = curA.next;
        curB = curB.next;
    }
    return curA;
}