Comments (2)
解题思路
- 遍历两链表, 比较大小后依次将其合并到主链表.
代码
public static class ListNode {
int val; // 当前链表值
ListNode next; // 下一个链表对象
ListNode(int x) { //赋值链表的值
val = x;
}
}
public static ListNode func(ListNode node1, ListNode node2) {
ListNode result = new ListNode(0);
ListNode tmp = result;
// 循环两个链表, 直到其中一个到底
while (node1 != null && node2 != null) {
// 每次选择val最小的node, 作为返回的node
if (node1.val > node2.val) {
ListNode swap = node1;
node1 = node2;
node2 = swap;
}
tmp.next = node1;
node1 = node1.next;
tmp = tmp.next;
}
// 获得未到底的链表, 继续赋值
tmp.next = node1 != null ? node1 : node2 ;
return result.next;
}
from pisecespeng.record.me.
LeetCode题解
解题思路
- 递归, 比较大小后依次将其合并到链表
代码
public static class ListNode {
int val; // 当前链表值
ListNode next; // 下一个链表对象
ListNode(int x) { //赋值链表的值
val = x;
}
}
public static ListNode func(ListNode node1, ListNode node2) {
if (node1 == null) {
return node2;
} else if (node2 == null) {
return node1;
} else if (node1.val < node2.val) {
node1.next = func(node1.next, node2);
return node1;
} else {
node2.next = func(node1, node2.next);
return node2;
}
}
from pisecespeng.record.me.
Related Issues (20)
- 两数相加 HOT 2
- 买卖股票的最佳时机 HOT 3
- 买卖股票的最佳时期2 HOT 3
- 二叉树的层序遍历 HOT 2
- 二叉树的最大深度 HOT 2
- 二进制求和 HOT 4
- 从尾到头打印链表 HOT 2
- 位1的个数 HOT 4
- 删除排序数组中的重复项 HOT 2
- 删除链表的倒数第N个节点 HOT 3
- 加一 HOT 2
- 反转字符串 HOT 2
- 反转链表 HOT 3
- 只出现一次的数字 HOT 2
- 合并两个有序数组 HOT 3
- 回文数 HOT 2
- 回文链表 HOT 3
- 复制带随机指针的链表 HOT 2
- 外观数列 HOT 3
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