力扣基础链表题总结

Posted 2021-09-23 Billy Miracle

237. 删除链表中的节点

思路：
将所给结点值改变，再将结点next指向下下个结点，就完成了要求。

One Possible Answer：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
void deleteNode(struct ListNode* node) {
    node->val = node->next->val;
    node->next = node->next->next;
}

19. 删除链表的倒数第 N 个结点

思路：
双指针，找到结点

One Possible Answer：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
    if (head == NULL) {
        return head;
    }
    struct ListNode* newhead = (struct ListNode*)malloc(sizeof(struct ListNode));
    newhead->val = 0, newhead->next = head;
    struct ListNode* first = head, * second = newhead;
    for (int i = 0; i < n; ++i) {
        first = first->next;
    }
    while (first) {
        first = first->next;
        second = second->next;
    }
    second->next = second->next->next;
    struct ListNode* ans = newhead->next;
    free(newhead);
    return ans;
}

206. 反转链表

思路：
从头一个个逆向建链，最后返回新链的头

One Possible Answer：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
typedef struct ListNode Node;
struct ListNode* reverseList(struct ListNode* head) {
	Node* pre = NULL, * cur = head;
	while (cur) {
		Node* next = cur->next;
		cur->next = pre;
		pre = cur;
		cur = next;
	}
    return pre;
}

21. 合并两个有序链表

思路：
建新链，一个个判断，把小的接在后面。

One Possible Answer：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    if (l1 == NULL) {
        return l2;
    }
    if (l2 == NULL) {
        return l1;
    }
	struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode)), * t=head;
	while (l1 && l2) {
		if (l1->val < l2->val) {
			t->next = l1;
			l1 = l1->next;
		} else {
			t->next = l2;
			l2 = l2->next;
		}
		t = t->next;
	}
	if (l1) {
		t->next = l1;
	} else if (l2) {
		t->next = l2;
	}
	return head->next;
}

234. 回文链表

思路：
把数字取到数组中再判断

One Possible Answer：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
bool isPalindrome(struct ListNode* head) {
    int vals[100000], l = 0;
    while (head != NULL) {
        vals[l++] = head->val;
        head = head->next;
    }
    for (int i = 0, j = l - 1; i < j; ++i, --j) {
        if (vals[i] != vals[j]) {
            return false;
        }
    }
    return true;
}

141. 环形链表

思路：
快慢指针

One Possible Answer：

bool hasCycle(struct ListNode* head) {
    if (head == NULL || head->next == NULL) {
        return false;
    }
    struct ListNode* slow = head;
    struct ListNode* fast = head->next;
    while (slow != fast) {
        if (fast == NULL || fast->next == NULL) {
            return false;
        }
        slow = slow->next;
        fast = fast->next->next;
    }
    return true;
}