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
     
      & 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
      
        result;12         int count;13         for(i=0;i
       
        2)25             {26                 res = res-(count-2);27                 result.push_back(temp);28                 result.push_back(temp);29                 30             }31             else32                 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
     
      & nums, int target) { 4         int n=nums.size(); 5         vector
      
        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                 else25                 left = midle + 1;26             }27             else {28                 if(A[midle] < target && target <= A[right])29                     left = midle + 1;30                 else31                     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             else37             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 Solution10 {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 };