剑指offer（16）链表合并

（19.1.11）

剑指offer（16）链表合并

题目描述
输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

第一想到的就是递归：

pHead1和pHead2为两者的头指针，比较它的pHead1.val和pHead2.val若pHead1.val小，就可以递归pHead1.next = Merge（pHead1.next, pHead2.next）反之， pHead2.next = Merg(pHead1,pHead2.next);

代码如下：

function ListNode(x){
    this.val = x;
    this.next = null;
}
function Merge(pHead1, pHead2)
{
    // write code here
    if(pHead1 == null) {
        return pHead2;
    }
    if(pHead2 == null) {
        return pHead1;
    }
    if(pHead1.val <= pHead2.val) {
        pHead1.next = Merge(pHead1.next, pHead2);
        return pHead1;
    }
    else {
        pHead2.next = Merge(pHead1,pHead2.next);
        return pHead2;
    }
}

非递归也是这个思想：

先创建一个root=head = new ListNode（-1）的根节点，之后while遍历pHead1和pHead2为非空，若pHead1.val<pHead2.val，则pHead1当前节点插入根节点head.next上，之后pHead1 = pHead1.next.反之，也差不多。

代码如下：

function MergeRecursion(pHead1, pHead2)
{
    // write code here
    var head = new ListNode(-1);
    var root = head;
    head.next = null;
    while(pHead1 !== null && pHead2 !== null) {
        if(pHead1.val <= pHead2.val) {
            head.next = pHead1;
            head = pHead1;
            pHead1 = pHead1.next;
        }
        else {
            head.next = pHead2;
            head = pHead2;
            pHead2 = pHead2.next;            
        }
    }
    if(pHead1 != null) {
        head.next = pHead1;
    }
    if(pHead2 != null) {
        head.next = pHead2;
    }
    return root.next;
}