143. 重排链表
cpp
复制代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* findMiddle(ListNode* head){
ListNode* p = head, *q = head;
// q移动2步，p移动一步
while(q->next && q->next->next){
q = q->next->next;
p = p->next;
}
return p;
}
ListNode* reverseList(ListNode* head){
ListNode* pre = nullptr, *cur = head, *next = nullptr;
while(cur){
next = cur->next;
cur->next = pre;
pre = cur;
cur = next;
}
return pre;
}
void mergeList(ListNode* l, ListNode* r){
ListNode* p = l, *q = r;
while(p && q){
ListNode* t1 = p->next, *t2 = q->next;
p->next = q;
p = t1;
q->next = p;
q = t2;
}
}
void reorderList(ListNode* head) {
if(head == nullptr) return;
// 思路：找到链表中点，右链表逆转，逐一合并左右链表节点
// 1. 找中点
ListNode* mid_head = findMiddle(head);
ListNode* right_head = mid_head->next;
mid_head->next = nullptr;
// 2. 右边翻转
right_head = reverseList(right_head);
// 3. 合并两个链表
mergeList(head, right_head);
}
};
2. 两数相加
- 2. 两数相加
- 思路：注释
- 时间：O(max(m, n))；空间：O(1)
cpp
复制代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
// 思路1：读取数字，倒序，相加，倒序，链表
// 思路2：对应位置相加，进位到后一位
ListNode* dummy_head = new ListNode();
ListNode* cur = dummy_head;
int carry = 0;
while(l1 || l2 || carry){
int a = l1 ? l1->val : 0;
int b = l2 ? l2->val : 0;
int sum = a + b + carry;
carry = sum / 10;
cur->next = new ListNode(sum % 10);
cur = cur -> next;
if(l1) l1 = l1->next;
if(l2) l2 = l2->next;
}
return dummy_head->next;
}
};
445. 两数相加 II
cpp
复制代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head){
ListNode* pre = nullptr, *cur = head, *next = nullptr;
while(cur){
next = cur->next;
cur->next = pre;
pre = cur;
cur = next;
}
return pre;
}
ListNode* addList(ListNode* lhs, ListNode* rhs){
ListNode* dummy_head = new ListNode();
ListNode* cur = dummy_head;
int carry = 0;
while(carry || lhs || rhs){
int a = lhs ? lhs->val : 0;
int b = rhs ? rhs->val : 0;
int sum = a + b + carry;
carry = sum / 10;
cur->next = new ListNode(sum % 10);
cur = cur->next;
if(lhs) lhs = lhs->next;
if(rhs) rhs = rhs->next;
}
return dummy_head->next;
}
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
// 反转2个链表，然后相加
ListNode* p = reverseList(l1), *q = reverseList(l2);
ListNode* ret = addList(p, q);
return reverseList(ret);
}
};