160. Intersecting Linked List | Algorithm (leetcode, with mind map + all solutions) 300 questions

Zero title: Algorithm (leetcode, with mind map + all solutions) (160) intersecting linked list of 300 questions

a topic description

Two solutions overview (mind map)

All three solutions

1 Scenario 1

1) Code:

 // 方案1 “自己。哈希法（JS里的Map数据结构）”。

// 思路：
// 1）状态初始化：resMap = new Map(), resNode = null; 。
// 2）核心1：遍历 链表A ，将每个节点存入 哈希resMap 中。
// 3）核心2：遍历 链表B 。
// 3.1）若 当前节点是否存在于 哈希resMap 中，则 resNode 置为当前节点 并 退出遍历。
// 4）返回结果 resNode 。
var getIntersectionNode = function(headA, headB) {
    // 1）状态初始化：resMap = new Map(), resNode = null; 。
    let resMap = new Map(),
        resNode = null;

    // 2）核心1：遍历 链表A ，将每个节点存入 哈希resMap 中。
    while (headA) {
        resMap.set(headA, 1);
        headA = headA.next;
    }

    // 3）核心2：遍历 链表B 。
    while (headB) {
        // 3.1）若 当前节点是否存在于 哈希resMap 中，则 resNode 置为当前节点 并 退出遍历。
        if (resMap.has(headB)) {
            resNode = headB;
            break;
        }
        headB = headB.next;
    }

    // 4）返回结果 resNode 。
    return resNode;
}

2 Option 2

1) Code:

 // 方案2 “双指针法。”。
// 参考：
// 1）https://leetcode.cn/problems/intersection-of-two-linked-lists/solution/xiang-jiao-lian-biao-by-leetcode-solutio-a8jn/

// 思路：
// 1）每步操作需要同时更新指针 pA 和 pB。
// 2）若 指针 pA 不为空，则 将指针 pA 移到下一个节点；若 指针 pB 不为空，则 将指针 pB 移到下一个节点。
// 3）若 指针 pA 为空，则 将指针 pA 移到链表 headB 的头节点；若 指针 pB 为空，则 将指针 pB 移到链表 headA 的头节点。
// 4）若 当指针 pA 和 pB 指向同一个节点 或 都为空时，则 返回它们指向的节点 或 null 。
var getIntersectionNode = function(headA, headB) {
    let pA = headA, pB = headB;

    while (pA !== pB) {
        pA = pA === null ? headB : pA.next;
        pB = pB === null ? headA : pB.next;
    }

    return pA;
}

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