leetcode Merge Two Sorted Lists(Java)
2017-06-24 22:08
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题目链接:点击打开链接
类型:合并
解法:递归
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;
if (l1.val < l2.val)
{
l1.next = mergeTwoLists(l1.next, l2);
return l1;
}
else
{
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
}
解法:遍历
ListNode resNode = new ListNode(0);
ListNode curNode = resNode;
while (l1 != null && l2 != null)
{
if (l1.val <= l2.val)
{
curNode.next = l1;
l1 = l1.next;
}
else
{
curNode.next = l2;
l2 = l2.next;
}
curNode = curNode.next;
}
curNode.next = (l1==null)?l2:l1;
return resNode.next;
类型:合并
解法:递归
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;
if (l1.val < l2.val)
{
l1.next = mergeTwoLists(l1.next, l2);
return l1;
}
else
{
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
}
解法:遍历
ListNode resNode = new ListNode(0);
ListNode curNode = resNode;
while (l1 != null && l2 != null)
{
if (l1.val <= l2.val)
{
curNode.next = l1;
l1 = l1.next;
}
else
{
curNode.next = l2;
l2 = l2.next;
}
curNode = curNode.next;
}
curNode.next = (l1==null)?l2:l1;
return resNode.next;
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