LeetCode开心刷题三十八天——83. Remove Duplicates from Sorted List88. Merge Sorted Array

83. Remove Duplicates from Sorted List

Easy

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Given a sorted linked list, delete all duplicates such that each element appear only once.

Example 1:

Input: 1->1->2
Output: 1->2

Example 2:

Input: 1->1->2->3->3
Output: 1->2->3

BONUS：

很多时候，在对结构体进行相应的编码时，时而发现是用点运算符( . )，时而是用箭头运算符( -> )；那么这两者之间的使用有什么区别么？

相同点：两者都是二元操作符，而且右边的操作数都是成员的名称。
不同点：点运算符( . )的左边操作数是一个结果为结构的表达式；
              箭头运算符( -> )的左边的操作数是一个指向结构体的指针。

例如：

typedef struct          // 定义一个结构体类型：DATA
{
    char key[10];       // 结构体成员：key
    char name[20];      // 结构体成员：name
    int age;            // 结构体成员：age
}DATA;
    
DATA data;              // 声明一个结构体变量
DATA *pdata;            // 声明一个指向结构体的指针
    
// 访问数据操作如下：
data.age = 24;          // 结构体变量通过点运算符( . )访问
pdata->age = 24;        // 指向结构体的指针通过箭头运算符( -> )访问

 

 Python version：
method 1:recursion seems the simplest method.
short:slow

# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
　　　　　#or need two not before and behind 
        if not head or not head.next: return head
        head.next=self.deleteDuplicates(head.next)
        return head if head.val !=head.next.val else head.next


solu=Solution()
node1=ListNode(1)
node2=ListNode(2)
node3=ListNode(2)
node4=ListNode(4)
node1.next=node2
node2.next=node3
node3.next=node4
head=node1
solu.deleteDuplicates(head)

while head:
    print(head.val)
    head=head.next

 

88. Merge Sorted Array

Easy

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Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

• The number of elements initialized in nums1 and nums2 are m and n respectively.
• You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.

Example:

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3

Output: [1,2,2,3,5,6]
ATTENTION：
1.when print nums1 directly put judge condition is <tail,I just thought tail is the 
size of nums1,but tail-- had lost it size,so cannot directly use at this part 
2.nums1 is the full-length one ,it's large enough to store all nums1&nums2,so this needn't
return is a void function.so use two index n+m-1.n-1 represent different part of nums1 to
make sum

#include<iostream>
#include<string>
#include<vector>
using namespace std;
class Solution
{
public:
    void merge(vector<int>& nums1,int m,vector<int>& nums2,int n)
    {
        int i=m-1;
        int j=n-1;
        int tail=n+m-1;
        while(j>=0)
        {
            nums1[tail--]=(i>=0&&nums1[i]>nums2[j])?nums1[i--]:nums2[j--];
        }
        cout<<"END"<<endl;
        for(int i=0;i<n+m;i++)
        {
            cout<<nums1[i]<<endl;
        }
    }
};
int main()
{
    Solution s1;
    vector<int> nums1;
    nums1.push_back(1);
    nums1.push_back(2);nums1.push_back(3);nums1.push_back(0);nums1.push_back(0);nums1.push_back(0);
    cout<<"flag"<<endl;
    for(int i=0;i<nums1.size();i++)
    {
        cout<<nums1[i]<<endl;
    }
    vector<int> nums2;nums2.push_back(4);nums2.push_back(5);nums2.push_back(6);
    for(int i=0;i<nums2.size();i++)
    {
        cout<<nums2[i]<<endl;
    }
    int m=3,n=3;
    s1.merge(nums1,m,nums2,n);
    return 0;
}