如何求列表的sort？就是用歸并排序，好好學(xué)習(xí)這個(gè)方法，遞歸先求中點(diǎn)，然后求前面的sort，之后求后面的sort，最后merge！
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* merge(ListNode* h1, ListNode* h2){ ListNode* hn = new ListNode(INT_MIN); ListNode* c = hn; while(h1 != NULL && h2 != NULL){ if(h1 -> val < h2 -> val){ c -> next = h1; h1 = h1 -> next; } else{ c -> next = h2; h2 = h2 -> next; } c = c -> next; } if(h1 != NULL) c -> next = h1; if(h2 != NULL) c -> next = h2; return hn -> next; } ListNode* sortList(ListNode* head) { if(head == NULL || head -> next == NULL) return head; ListNode* f = head -> next -> next; ListNode* p = head; while(f != NULL && f -> next != NULL){ p = p -> next; f = f -> next -> next; } ListNode* h = sortList(p -> next); p -> next = NULL; return merge(sortList(head), h); }};新聞熱點(diǎn)
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