题解 | #合并两个排序的链表#同时遍历选最小点/递归/插入

解法1: 遍历两个链表，每次选最小的元素作为下一个元素

public class Solution {
    public ListNode Merge(ListNode list1, ListNode list2) {
        if (list1 == null) {
            return list2;
        }
        if (list2 == null) {
            return list1;
        }

        ListNode newHead;
        if (list1.val <= list2.val) {
            newHead = list1;
            list1 = list1.next;
        } else {
            newHead = list2;
            list2 = list2.next;
        }
        ListNode prev = newHead;
        //iterate two lists, select node with smaller val to be the next node;
        while (list1 != null && list2 != null) {
            // if list1 is smaller, then next node is list1, and move pointer forward
            if (list1.val < list2.val) {
                prev.next = list1;
                prev = list1;
                list1 = list1.next;
            } else {
                prev.next = list2;
                prev = list2;
                list2 = list2.next;
            }

        }
        // after the iterator, must have one list not empty, add it to the end of the result
        if (list1 == null) {
            prev.next = list2;
        } else {
            prev.next = list1;
        }
        return newHead;
    }
}

解法2: 递归

public class Solution {
    public ListNode Merge(ListNode list1, ListNode list2) {
        if (list1 == null) {
            return list2;
        }
        if (list2 == null) {
            return list1;
        }

        ListNode newHead;
        if (list1.val <= list2.val) {
            newHead = list1;
            newHead.next = Merge(list1.next, list2);
        } else {
            newHead = list2;
            newHead.next = Merge(list1, list2.next);
        }
        return newHead;
    }
}

解法3: 将链表2插入到链表1中

public class Solution {
    public ListNode Merge(ListNode list1, ListNode list2) {
        if (list1 == null) {
            return list2;
        }
        if (list2 == null) {
            return list1;
        }

        ListNode newHead;
        if (list1.val < list2.val) {
            newHead = list1;
        } else {
            newHead = list2;
        }
        ListNode prevOfInsert = new ListNode(0x80000000);
        prevOfInsert.next = list1;
        ListNode nextOfInsert = list1;
        ListNode toBeInsert;
        while (list2 != null && nextOfInsert != null) {
            toBeInsert = list2;
            if (list2.val <= nextOfInsert.val) {
                list2 = list2.next;
                prevOfInsert.next = toBeInsert;
                toBeInsert.next = nextOfInsert;
                prevOfInsert = toBeInsert;
            } else {
                nextOfInsert = nextOfInsert.next;
                prevOfInsert = prevOfInsert.next;
            }
        }
        // if there are nodes in list2 that greater than the last node of list1, append all list2 to the end
        if (nextOfInsert == null && list2 != null) {
            prevOfInsert.next = list2;
        }
        return newHead;
    }
}