80. Remove Duplicates from Sorted Array II
题目
分析：简单的操作，代码如下：
1 class Solution { 2 public: 3 int removeDuplicates(vector<int>& nums) { 4 int n = nums.size(); 5 if(0==n) 6 return 0; 7 8 int i=0; 9 int temp; 10 int res = n; 11 vector<int> result; 12 int count; 13 for(i=0;i<n;) 14 { 15 temp=nums[i]; 16 count=1; 17 i++; 18 while(i<n&&nums[i] == temp) 19 { 20 count++; 21 i++; 22 } 23 24 if(count>2) 25 { 26 res = res-(count-2); 27 result.push_back(temp); 28 result.push_back(temp); 29 30 } 31 else 32 while(count--) 33 { 34 result.push_back(temp); 35 } 36 37 } 38 nums = result; 39 return res; 40 41 } 42 };
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81. Search in Rotated Sorted Array II
题目
分析：题目和33题很相识，代码如下：
1 class Solution { 2 public: 3 bool search(vector<int>& nums, int target) { 4 int n=nums.size(); 5 vector<int> A=nums; 6 if(0 == n) return false; 7 int left = 0; 8 int right = n - 1; 9 while(left <= right) 10 { 11 int midle = (left + right) >> 1; 12 if(A[midle] == target) return true; 13 if(A[left] == A[midle] && A[midle] == A[right]) 14 { 15 ++left; 16 --right; 17 } 18 else if(A[left] <= A[midle]) 19 { 20 if(A[left] <= target && target < A[midle]) 21 { 22 right = midle - 1; 23 } 24 else 25 left = midle + 1; 26 } 27 else { 28 if(A[midle] < target && target <= A[right]) 29 left = midle + 1; 30 else 31 right = midle - 1; 32 } 33 } 34 return false; 35 } 36 };
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82. Remove Duplicates from Sorted List II
题目
分析：这道题主要是考察指针操作，为了方便，在处理之前，对链表添加一个头节点，以便处理起来更加方便，代码如下
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(NULL == head) 13 return NULL; 14 15 ListNode *pre,*current; 16 ListNode *pHead = new ListNode(0); 17 pHead->next = head;//添加头节点 18 pre = pHead; 19 current = head; 20 int key; 21 bool flag = false; 22 while(current!= NULL) 23 { 24 key = current->val; 25 current = current->next; 26 while( current != NULL && current->val == key) 27 { 28 flag = true; 29 current = current->next; 30 } 31 if(flag) 32 { 33 pre->next = current; 34 flag = false; 35 } 36 else 37 pre = pre->next; 38 } 39 40 return pHead->next; 41 42 } 43 };
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83. Remove Duplicates from Sorted List
分析：这一题和82题类似，代码如下：
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 { 11 public: 12 ListNode *deleteDuplicates(ListNode *head) 13 { 14 if(head==NULL || head->next==NULL) return head; 15 ListNode *helper = new ListNode(-100000); 16 ListNode *ret=head; 17 while(ret) 18 { 19 ListNode *next=ret->next; 20 if(ret->val!=helper->val) 21 { 22 helper->next=ret; 23 helper=ret;//将helper指新链表的尾结点 24 helper->next=NULL;//尾指向空，因为后面的结点有可能被删去了，它不知道下一个指向谁 25 } 26 else delete ret; 27 ret=next; 28 } 29 return head; 30 } 31 };