一、移除链表元素
示例 1:
输入:head = [1,2,6,3,4,5,6], val = 6 输出:[1,2,3,4,5]
示例 2:
1. 输入:head = [], val = 1 2. 输出:[]
示例 3:
1. 输入:head = [7,7,7,7], val = 7 2. 输出:[]
提示:
- 列表中的节点数目在范围
[0, 104]
内1 <= Node.val <= 50
0 <= val <= 50
方法一:
代码解析:
struct ListNode* removeElements(struct ListNode* head, int val){ struct ListNode* prev = NULL; struct ListNode* cur = head; while(cur) { if(cur->val != val) { prev = cur; cur = cur->next; } else { if(prev == NULL) { head = cur->next; free(cur); cur = head; } else { prev->next = cur->next; free(cur); cur = prev->next; } } } return head; }
画图解析:
方法二:
代码解析:
struct ListNode* removeElements(struct ListNode* head, int val){ if(head == NULL) { return NULL; } struct ListNode* newHead = NULL,*tail = NULL; struct ListNode* cur = head; while(cur) { if(cur->val != val) { //尾插 if(tail == NULL) { newHead = tail = cur; } else { tail->next =cur; tail = tail->next; } cur = cur->next; } else { struct ListNode* next = cur->next; free(cur); cur = next; } } if(tail) tail->next = NULL; return newHead; }
画图解析:
二、分享:OJ调试技巧
LeetCode在线调试功能需要付费,我们可以自己在编辑器进行调试,写个模板,每次用到复制过去直接调试即可
int main() { struct ListNode* n1 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* n2 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* n3 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* n4 = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* n5 = (struct ListNode*)malloc(sizeof(struct ListNode)); n1->val = 7; n2->val = 7; n3->val = 7; n4->val = 7; n5->val = 7; n1->next = n2; n2->next = n3; n3->next = n4; n4->next = NULL; removeElements(n1,7); }
三、链表的中间结点
示例 1:
输入:head = [1,2,3,4,5] 输出:[3,4,5] 解释:链表只有一个中间结点,值为 3 。
示例 2:
输入:head = [1,2,3,4,5,6] 输出:[4,5,6] 解释:该链表有两个中间结点,值分别为 3 和 4 ,返回第二个结点。
提示:
- 链表的结点数范围是
[1, 100]
1 <= Node.val <= 100
代码解析:
此题我们学会了快慢指针
struct ListNode* middleNode(struct ListNode* head){ struct ListNode* slow,* fast; slow = fast = head; while(fast && fast->next)//节点为单数或者双数时出现的条件 { slow = slow->next; fast = fast->next->next; } return slow; }
画图解析:
四、链表中倒数第k个结点
输入一个链表,输出该链表中倒数第k个结点。
实例:
输入:1,{1,2,3,4,5} 返回值:{5}
代码解析:
此题依旧使用快慢指针
struct ListNode* FindKthToTail(struct ListNode* pListHead, int k ) { if(pListHead == NULL) { return NULL; } struct ListNode* slow,* fast; slow = fast = pListHead; //fast while(k--) { if(fast == NULL) { return NULL; } fast = fast->next; } while(fast) { slow = slow->next; fast = fast->next; } return slow; }
画图解析: