标签：
138题：Copy List with Random Pointer
题目分析：
本题思路1：第一步，你需要遍历一下链表，对于每个结点，你都new出一个连接在其后面。第二步，调整random指针。第三步，把复制的链表与原链表断开。时间复杂度O(N)，空间复杂度O(1)。
本题思路2：第一步，仍需要遍历一下链表，对于每个结点都new出一个节点，但不连接在其后面，把这种旧节点到新结点的映射关系，存储在map中。第二步，调整random指针。时间复杂度O(N)，空间复杂度O(N)。
本题思路3：第一步与前两个思路相同，但不记录random信息。第二步，对于new出的每个结点，遍历原链表一次，找到合适位置。时间复杂度O(N^2)，空间复杂度O(1)。
思路1解法：
1 class Solution { 2 public: 3 RandomListNode* copyRandomList(RandomListNode *head) 4 { 5 if (head == NULL) 6 { 7 return NULL; 8 } 9 10 //copy Nodes and ignore random pointers 11 RandomListNode *cur = head, *cpycur = NULL; 12 while (cur != NULL) 13 { 14 cpycur = new RandomListNode(cur->label); 15 cpycur->next = cur->next; 16 cur->next = cpycur; 17 cur = cpycur->next; 18 } 19 20 //Adjust the random pointers 21 for (cur = head, cpycur = cur->next; cur; cur = cpycur->next, cpycur = cur != NULL ? cur->next : NULL) 22 { 23 if (cur->random != NULL) 24 { 25 cpycur->random = cur->random->next; 26 } 27 } 28 29 //separate the list 30 cur = head; 31 RandomListNode *cpyhead = cur->next; 32 for (cpycur = cpyhead; cur; ) 33 { 34 cur->next = cpycur->next; 35 cur = cur->next; 36 cpycur->next = cur != NULL ? cur->next : NULL; 37 cpycur = cpycur->next; 38 } 39 40 return cpyhead; 41 } 42 };
思路2解法：
1 class Solution { 2 public: 3 RandomListNode* copyRandomList(RandomListNode *head) 4 { 5 if (head == NULL) 6 { 7 return NULL; 8 } 9 10 //copy Nodes and ignore random pointers 11 RandomListNode *cur = head; 12 RandomListNode dummy(-1); 13 RandomListNode *cpypre = &dummy, *cpycur = dummy.next; 14 15 unordered_map<RandomListNode*, RandomListNode*> cpymap; 16 while (cur != NULL) 17 { 18 cpycur = new RandomListNode(cur->label); 19 cpypre->next = cpycur; 20 cpymap[cur] = cpycur; 21 cur = cur->next; 22 cpypre = cpypre->next; 23 } 24 25 //Adjust random pointers 26 for (cur = head; cur; cur = cur->next) 27 { 28 cpymap[cur]->random = cpymap[cur->random]; 29 } 30 31 return dummy.next; 32 } 33 };
思路3就不提供了，这个，没什么太大价值。
109题：Convert Sorted List to Binary Search Tree
题目分析：递归将链表中点作为子树的根即可。下面解法时间复杂度O(N^2)，空间复杂度O(logN)。（好吧，其实我也不知道空间复杂度为啥是logN）
1 class Solution { 2 public: 3 ListNode* FindnthPos(ListNode *head, int n) 4 { 5 if (n <= 0) 6 { 7 return NULL; 8 } 9 10 ListNode *p = head; 11 while (--n) 12 { 13 p = p->next; 14 } 15 return p; 16 } 17 18 TreeNode* PutsortedListToBST(ListNode *head, int len) 19 { 20 if (len == 0) 21 { 22 return NULL; 23 } 24 if (len == 1) 25 { 26 return new TreeNode(head->val); 27 } 28 29 TreeNode *root = new TreeNode(FindnthPos(head, len / 2 + 1)->val); 30 root->left = PutsortedListToBST(head, len / 2); 31 root->right = PutsortedListToBST(FindnthPos(head, len / 2 + 2), (len - 1) / 2); 32 33 return root; 34 } 35 36 TreeNode* sortedListToBST(ListNode *head) 37 { 38 int len = 0; 39 ListNode *p = head; 40 while (p != NULL) 41 { 42 len = len + 1; 43 p = p->next; 44 } 45 46 return PutsortedListToBST(head, len); 47 } 48 };
另外一种自底向上的解法，时间复杂度能够达到O(N)，空间复杂度O(logN)。
1 class Solution { 2 public: 3 TreeNode *sortedListToBST(ListNode *head) 4 { 5 int len = 0; 6 ListNode *p = head; 7 while (p) { 8 len++; 9 p = p->next; 10 } 11 return sortedListToBST(head, 0, len - 1); 12 } 13 private: 14 TreeNode* sortedListToBST(ListNode*& list, int start, int end) 15 { 16 if (start > end) return nullptr; 17 int mid = start + (end - start) / 2; 18 TreeNode *leftChild = sortedListToBST(list, start, mid - 1); 19 TreeNode *parent = new TreeNode(list->val); 20 parent->left = leftChild; 21 list = list->next; 22 parent->right = sortedListToBST(list, mid + 1, end); 23 return parent; 24 } 25 };
191题：Number of 1 Bits
题目分析：数1的个数。这个我貌似在程序员面试宝典里看过，给出一个程序，问干啥的，答：数1的个数。哈哈…
1 class Solution { 2 public: 3 int hammingWeight(uint32_t n) 4 { 5 int count = 0; 6 while (n) 7 { 8 count = count + 1; 9 n = n & (n - 1); 10 } 11 return count; 12 } 13 };
________________________________________________我是分割线___________________________________________
LeetCode OJ 其它题目：
LeetCode OJ Linked List: 24题、148题和61题
LeetCode OJ Linked List: 92题、143题、147题和160题
LeetCode OJ Linked List: 19题、83题、82题和86题
LeetCode OJ Linked List: 21题、2题、141题和142题
LeetCode OJ Linked List: 138题、109题和191题
标签：
原文地址：http://www.cnblogs.com/mengwang024/p/4369838.html
