LeetCode之Merge In Between Linked

问题：

方法：
第一步，遍历list1，通过序号获取a对应的节点和b对应的节点；第二步，遍历list2获取尾部节点，然后把list2的头部接到a节点后，把b节点接到list2的尾部，最后返回list1的头部节点即可。

class MergeInBetweenLinkedLists {
    fun mergeInBetween(list1: ListNode?, a: Int, b: Int, list2: ListNode?): ListNode? {
        var aNode: ListNode? = null
        var bNode: ListNode? = null
        var index = 0
        var cur = list1
        while (cur != null) {
            if (index == a - 1) {
                aNode = cur
            }
            if (index == b + 1) {
                bNode = cur
            }
            cur = cur.next
            index++
        }
        aNode?.next = list2
        cur = list2
        while (cur?.next != null) {
            cur = cur.next
        }
        cur?.next = bNode
        return list1
    }
}

fun main() {

}

有问题随时沟通

具体代码实现可以参考Github