标签：style blog http io ar color os sp for
利用插入排序，对链表进行排序。
插入排序点here。
居然初次想用unordered map来存每个节点，然后根据下标就可以访问了，再更新下班，也就是n方的方法了。但是不知为什么超时了。插入排序不就是n方吗。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *insertionSortList(ListNode *head) { if (!head || !head -> next) return head; unordered_map<int, ListNode *> umap; ListNode *tmp = head; int cnt = 0; while(tmp) { umap[cnt++] = tmp; tmp = tmp -> next; } ListNode *pre = new ListNode(0); pre -> next = head; umap[-1] = pre; for (int i = 1; i < cnt; i++) { tmp = umap[i]; bool flag = false; int j = i - 1; while(j >= 0 && umap[i] -> val < umap[j] -> val) { j--; flag = true; } if (flag) {// 处理umap的下标 for (int k = i - 1; k > j; k--) umap[k+1] = umap[k]; umap[j+1] = tmp; // 处理指针指向 umap[i] -> next = umap[j+1] -> next; umap[j+1] -> next = umap[j] -> next; umap[j] -> next = umap[j+1]; } } head = pre -> next; delete pre; return head; } };
然后就不从后往前插入，而是从前往后判断，复杂度应该是一样的，只是不要用到map，避免了更新map下标的操作。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *insertionSortList(ListNode *head) { if (!head || !head -> next) return head; ListNode *pre = new ListNode(0); pre -> next = head; ListNode *p = head->next, *prep = head, *start = head, *preStart = pre; while(p) { // 每次从头往后找第一个比p大的 preStart = pre; start = preStart -> next;// 注意了每次的开始不是head，因为head可能已经变了，开始应该是pre的next while(start != p && start->val <= p->val) { preStart = preStart -> next; start = start->next; } if (start == p) { prep = p; p = p -> next; } else { prep -> next = p -> next; p -> next = start; preStart -> next = p; p = prep -> next; // 继续下一次的时候是prep的下一个了，也就是之前p的下一个 } } head = pre -> next; delete pre; return head; } };
标签：style blog http io ar color os sp for
原文地址：http://www.cnblogs.com/higerzhang/p/4166028.html
