Leetcode题解（26）

80. Remove Duplicates from Sorted Array II

题目

分析：简单的操作，代码如下：

class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int n = nums.size();
if(0==n)
return 0;

int i=0;
int temp;
int res = n;
vector<int> result;
int count;
for(i=0;i<n;)
{
temp=nums[i];
count=1;
i++;
while(i<n&&nums[i] == temp)
{
count++;
i++;
}

if(count>2)
{
res = res-(count-2);
result.push_back(temp);
result.push_back(temp);

}
else
while(count--)
{
result.push_back(temp);
}

}
nums = result;
return res;

}
};

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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：

class Solution {
public:
bool search(vector<int>& nums, int target) {
int n=nums.size();
vector<int> A=nums;
if(0 == n) return false;
int left = 0;
int right = n - 1;
while(left <= right)
{
int midle = (left + right) >> 1;
if(A[midle] == target) return true;
if(A[left] == A[midle] && A[midle] == A[right])
{
++left;
--right;
}
else if(A[left] <= A[midle])
{
if(A[left] <= target && target  < A[midle])
{
right = midle - 1;
}
else
left = midle + 1;
}
else {
if(A[midle] < target && target <= A[right])
left = midle + 1;
else
right = midle - 1;
}
}
return false;
}
};

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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

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
if(NULL == head)
return NULL;

ListNode *pre,*current;
ListNode *pHead = new ListNode(0);
pHead->next = head;//添加头节点
pre = pHead;
current = head;
int key;
bool flag = false;
while(current!= NULL)
{
key = current->val;
current = current->next;
while( current != NULL && current->val == key)
{
flag = true;
current = current->next;
}
if(flag)
{
pre->next = current;
flag = false;
}
else
pre = pre->next;
}

return pHead->next;

}
};

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83. Remove Duplicates from Sorted List

分析：这一题和82题类似，代码如下：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution
{
public:
ListNode *deleteDuplicates(ListNode *head)
{
if(head==NULL || head->next==NULL) return head;
ListNode *helper = new ListNode(-100000);
ListNode *ret=head;
while(ret)
{
ListNode *next=ret->next;
if(ret->val!=helper->val)
{
helper->next=ret;
helper=ret;//将helper指新链表的尾结点
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
}
else delete ret;
ret=next;
}
return head;
}
};