Remove Duplicates from Sorted List I&&II
Remove Duplicates from Sorted List
Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
这题思路很清楚，当前节点的值与前一节点的值相同，那么就直接删掉
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *deleteDuplicates(ListNode *head) { 12 if( !head ) return 0; 13 ListNode* pre = head; //前驱 14 ListNode* cur = pre->next; //当前 15 while( cur ) { 16 if( cur->val == pre->val ) { //若与前驱相同，直接删掉 17 ListNode* p = cur; 18 cur = cur->next; 19 pre->next = cur; 20 delete p; 21 } else { //不同直接下移 22 cur = cur->next; 23 pre = pre->next; 24 } 25 } 26 return head; 27 } 28 };
Remove Duplicates from Sorted List II
Total Accepted: 21273 Total Submissions: 85697My Submissions
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
凡是重复的数据都要删除，可以设个标记，如果发现有重复节点，那么将其值赋予标记上，然后凡是遇到标记值的节点直接删除
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *deleteDuplicates(ListNode *head) { 12 if( !head ) return 0; 13 const int INF = 0x3fffffff; //设置个不可能值 14 ListNode node(-1); //布置头节点，便于删除 15 node.next = head; 16 ListNode* pre = &node; 17 ListNode* cur = head; 18 int needDelete = INF; //便于逻辑判断 19 while( cur ) { 20 if( cur->val == needDelete ) { //若等于需被删除值，那么直接删除 21 ListNode* p = cur; 22 cur = cur->next; 23 pre->next = cur; 24 delete p; 25 } else if( cur->next && cur->val == cur->next->val ) { //若发现有重复元素，那么记录需要删除的值 26 needDelete = cur->val; 27 } else { //后移节点 28 cur = cur->next; 29 pre = pre->next; 30 } 31 } 32 return node.next; 33 } 34 };
这里的不可能值，如果可能的话，那么就需要加逻辑判断，否则就会出大bug，尤其是面试官询问的话
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *deleteDuplicates(ListNode *head) { 12 if( !head ) return 0; 13 ListNode node(-1); //布置头节点，便于删除 14 node.next = head; 15 ListNode* pre = &node; 16 ListNode* cur = head; 17 while( cur && cur->next && cur->val != cur->next->val ) { 18 pre = cur; 19 cur = cur->next; 20 } 21 if( !cur->next ) return node.next; 22 int needDelete = cur->val; 23 while( cur ) { 24 if( cur->val == needDelete ) { //若等于需被删除值，那么直接删除 25 ListNode* p = cur; 26 cur = cur->next; 27 pre->next = cur; 28 delete p; 29 } else if( cur->next && cur->val == cur->next->val ) { //若发现有重复元素，那么记录需要删除的值 30 needDelete = cur->val; 31 } else { //后移节点 32 cur = cur->next; 33 pre = pre->next; 34 } 35 } 36 return node.next; 37 } 38 };
