LeetCode刷题笔记-21.合并两个有序链表

C代码

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


struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2){
    struct ListNode *dummy = NULL;
    struct ListNode *head = NULL;

    if (list1 == NULL) {
        return list2;
    }

    if (list2 == NULL) {
        return list1;
    }

    if(list1->val <= list2->val) {
        dummy = list1;
        list1 = list1->next;
    } else {
        dummy = list2;
        list2 = list2->next;
    }
    head = dummy;

    while(list1 != NULL && list2 != NULL) {
        if (list1->val >= list2->val) {
            dummy->next = list2;
            list2 = list2->next;
        } else {
            dummy->next = list1;
            list1 = list1->next;
        }
        dummy = dummy->next;
    }
    if (list1 != NULL) {
        dummy->next = list1;
    }
    if (list2 != NULL) {
        dummy->next = list2;
    }
    return head;
}

注意点

  1. dummy和head节点的初始化,需要移动list1或者list2;
  2. 遍历操作感觉比较冗余(头初始化list1或者list2,后去遍历代码又移动一次),代码需要优化;

结果

LeetCode刷题笔记-21.合并两个有序链表_第1张图片

DUMMY C代码优化版本

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


struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2){
    struct ListNode *dummy = NULL;
    struct ListNode *head = NULL;

    if (list1 == NULL) {
        return list2;
    }

    if (list2 == NULL) {
        return list1;
    }

    head = calloc(1, sizeof(struct ListNode));
    dummy = head;

    while(list1 != NULL && list2 != NULL) {
        if (list1->val >= list2->val) {
            dummy->next = list2;
            list2 = list2->next;
        } else {
            dummy->next = list1;
            list1 = list1->next;
        }
        dummy = dummy->next;
    }
    if (list1 != NULL) {
        dummy->next = list1;
    }
    if (list2 != NULL) {
        dummy->next = list2;
    }
    dummy = head->next;
    free(head);
    head = NULL;
    return dummy;
}

思路

  1. 新建头节点空间head;
  2. 最终返回head->next,来统一list1与list2的移动逻辑到while循环内;

题目

LeetCode刷题笔记-21.合并两个有序链表_第2张图片

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