Leetcode题解（26）

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 };