Leetcode: Sort List - 归并

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Sort a linked list in O(n log n) time using constant space complexity.

以前写过快排的，很复杂，也容易出错。用归并试试，发现相当简单，逻辑也很清楚。一次通过。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *sortList(ListNode *head) {        if (head == NULL || head->next == NULL) {            return head;        }                ListNode *fast = head;        ListNode *slow = head;        ListNode *tail = slow;        while (fast != NULL && fast->next != NULL) {            fast = fast->next->next;            tail = slow;            slow = slow->next;        }                tail->next = NULL;        ListNode *first = sortList(head);        ListNode *second = sortList(slow);                // merge the 2 sorted lists        head = new ListNode(0);        tail = head;        while (first != NULL && second != NULL) {            if (first->val <= second->val) {                tail->next = first;                first = first->next;            }            else {                tail->next = second;                second = second->next;            }            tail = tail->next;        }                if (first != NULL) {            tail->next = first;        }        else if (second != NULL) {            tail->next = second;        }                tail = head->next;        delete head;        return tail;    }};

=======================第二次======================

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *sortList(ListNode *head) {        if (head == NULL || head->next == NULL) {            return head;        }                ListNode *slow = head;        ListNode *fast = head;        ListNode *prev = NULL;        while (fast != NULL && fast->next != NULL) {            fast = fast->next->next;            prev = slow;            slow = slow->next;        }        prev->next = NULL;                head = sortList(head);        slow = sortList(slow);                ListNode *new_head = new ListNode(0);        ListNode *cur = new_head;        while (head != NULL && slow != NULL) {            if (head->val <= slow->val)  {                cur->next = head;                head = head->next;            }            else {                cur->next = slow;                slow = slow->next;            }            cur = cur->next;        }                if (head != NULL) {            cur->next = head;        }        else if (slow != NULL) {            cur->next = slow;        }                head = new_head->next;        delete new_head;                return head;    }};

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