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K个排序链表归并【23】(hard) #include <iostream> #include <algorithm> #include <vector> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; bool cmp(const ListNode *a, const Lis

2个排序链表归并【21】(easy) #include <iostream> using namespace std; struct ListNode { int val; ListNode* next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {

复杂链表的复制【138】(medium) #include <iostream> #include <map> #include <vector> using namespace std; class Node { public: int val; Node *next; Node *random; Node(int _val) { val = _val; next = NULL;

链表划分【86】(medium) #include <iostream> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode *partition(ListNode *head, int x) { ListNode

链表求环【142】(medium) #include <iostream> #include <set> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode *detectCycle(ListNode *head) {

链表求交点 【160】(easy) #include <iostream> #include <set> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; int get_list_length(ListNode *head) { int len = 0; while (head

1-b. 链表逆序2 【92】(medium) #include <iostream> using namespace std; /* Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of list. Example: Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4-&

1-a. 链表逆序 【206】(easy) #include <stdio.h> struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode *reverseList(ListNode *head) { // 方法一、双指针迭代 ListNode *