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Muhammad al-Xorazmiy nomidagi Toshkent Axborot Texnologiyalar Universititeti “R va ma” fakulteti
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| Sana | 17.02.2023 | | Hajmi | 0.56 Mb. | | #42602 |
Bog'liq Alijonov.M MTA L-2 Alijonov.M MTA L-1, Lecture 5 - Polymorphism. Overloading
Muhammad al-Xorazmiy nomidagi
Toshkent Axborot Texnologiyalar
Universititeti “R va MA” fakulteti
051-20 guruh talabasi Alijonov Muhammadqodir.
Ma’lumotlar tuzilmasi va algoritmlar.
Kodlar C# (.NET 6) da yozildi.Masalalar LeetCode saytidan olindi.
Bajardi:Alijonov Muhammadqodir.
Tekshirdi:
Toshkent 2022
1-masala;
English version:
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string "".
UZB version:
Satrlar massivi orasidan eng uzun umumiy prefiks qatorini topish funksiyasini yozing.
Agar umumiy prefiks bo'lmasa, bo'sh "" qatorini qaytaring.
public class Solution {
public string LongestCommonPrefix(string[] strs) {
var result = new StringBuilder();
for(short i = 0; i < strs[0].Length; i++)
{
char c = strs[0][i];
for(short j = 1; j < strs.Length; j++)
if(strs[j].Length -1 < i || c != strs[j][i])
return result.ToString();
result.Append(c);
}
return result.ToString();
}
}
2-masala:
English version:
You are given the heads of two sorted linked lists list1 and list2.
Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.
Return the head of the merged linked list.
UZB version:
Sizga ikkita saralangan bog'langan ro'yxatlar ro'yxati 1 va ro'yxat 2 beriladi.
Ikki roʻyxatni bitta tartiblangan roʻyxatda birlashtiring. Ro'yxat birinchi ikkita ro'yxatning tugunlarini birlashtirish orqali tuzilishi kerak.Birlashtirilgan bog'langan ro'yxatning boshini qaytaring.
public class Solution {
public ListNode MergeTwoLists(ListNode list1, ListNode list2) {
return recursion(list1,list2);
}
public ListNode recursion(ListNode l1, ListNode l2)
{
if(l1 == null)
{
return l2;
}
if(l2==null)
{
return l1;
}
if(l1.val>l2.val)
{
return new ListNode(l2.val,recursion(l2.next, l1));
}
else
{
return new ListNode(l1.val, recursion(l2, l1.next));
}
}
}
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