剑指offer-合并两个排序的链表

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。

 其实思路跟归并排序差不多，注意边界值

    1.两个List都为空,则返回空
    2.其中一个List为空，则返回不为空的List

 循环解法：

public static ListNode merge(ListNode list1,ListNode list2) {
		if(list1==null||list2==null) {
			if(list1==null) {
				return list2;
			} else if (list2==null) {
				return list1;
			} else {
				return null;
			}
		}
		ListNode leftCurrent=list1;
		ListNode rightCurrent=list2;
		ListNode prehead=new ListNode(0);
		ListNode leftPrevious=null;
		ListNode rightPrevious=null;
		//先决定谁是头指针
		if(leftCurrent.val>rightCurrent.val) {
			prehead.next=rightCurrent;
			rightPrevious=rightCurrent;
			rightCurrent=rightCurrent.next;
		} else {
			prehead.next=leftCurrent;
			leftPrevious=leftCurrent;
			leftCurrent=leftCurrent.next;
		}
		//进行合并,当两者都没有到达最后一个点时
		while(rightCurrent!=null&&leftCurrent!=null) {
			if(leftCurrent.val>rightCurrent.val) {
				leftPrevious.next=rightCurrent;
				rightPrevious=rightCurrent;
				rightCurrent=rightCurrent.next;
			} else {
				rightPrevious.next=leftCurrent;
				leftPrevious=leftCurrent;
				leftCurrent=leftCurrent.next;
			}
		}
		//若右边链表不为空
		if(rightCurrent!=null) {
			leftPrevious.next=rightCurrent;
		}
		//若左边链表不为空
		if(leftCurrent!=null) {
			rightPrevious.next=leftCurrent;
		}
		return prehead.next;
}

　递归解法：

public static ListNode merge(ListNode list1,ListNode list2) {
		if(list1==null||list2==null) {
			if(list1==null) {
				return list2;
			} else if (list2==null) {
				return list1;
			} else {
				return null;
			}
		}
		if(list1.val>list2.val) {
			list2.next=merge(list1,list2.next);
			return list2;
		}  else {
			list1.next=merge(list1.next,list2);
			return list1;
		}
}