leetcode 21_Merge Two Sorted Lists & leetcode_258 Add Digits & leetcode_66plus one

最新推荐文章于 2025-12-06 07:43:37 发布

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#leetcode #MergeTwoSortedLists #Add Digits #plus one

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本文深入探讨了LeetCode中的两道经典算法题：如何合并两个有序链表，并给出了解决方案；以及如何通过重复对数字进行求和直至得到一位数，同时提供了具体的实现代码。

l leetcode 21_Merge Two Sorted Lists

题目：Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

解法：

#include <iostream>
using namespace std;

struct ListNode 
{
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};
 
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) 
    {
		if(l1 == NULL)
			return l2;
		if(l2 == NULL)
			return l1;
		ListNode *head = new ListNode(0);
		ListNode *p = head;
		while (l1 != NULL && l2 != NULL)
		{
			if (l1->val < l2->val)
			{
				ListNode *temp = new ListNode(l1->val);
				p->next = temp;
				p = temp;
				l1 = l1->next;
			}
			else
			{
				ListNode *temp = new ListNode(l2->val);
				p->next = temp;
				p = temp;
				l2 = l2->next;
			}
		}
        while (l1 != NULL)
        {
			ListNode *temp = new ListNode(l1->val);
			p->next = temp;
			p = temp;
			l1 = l1->next;
        }
		while (l2 != NULL)
		{
			ListNode *temp = new ListNode(l2->val);
			p->next = temp;
			p = temp;
			l2 = l2->next;
		}
		return head->next;
    }
};

leetcode_258 Add Digits.cpp

题目：

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

For example:

Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

Follow up:
Could you do it without any loop/recursion in O(1) runtime?

答案：

#include <iostream>
using namespace std;
class Solution 
{
public:
	int addDigits(int num) 
	{
		if(num < 10)
			return num;
		int tmp = 0;
		while(num >= 10)
		{
			tmp += num % 10;
			num = num / 10;
			if (num < 10 )
			{
				tmp += num;
				num = tmp;
				if (num < 10)
					return tmp;
				else
					tmp = 0;
			}
		}
	}
};

leetcode_66 plus one
题目：

Given a non-negative number represented as an array of digits, plus one to the number.

The digits are stored such that the most significant digit is at the head of the list.

解法：

#include <iostream>
#include <vector>
using namespace std;
class Solution 
{
public:
	vector<int> plusOne(vector<int>& digits) 
	{
		int length = digits.size();
		int carry;
		if (digits[length - 1] == 9)
			carry = 1;
		else
		{
			digits[length - 1] += 1;
			return digits;
		}
		for (int i = length - 1; i > 0; --i)
		{
			digits[i] += carry;
			if (digits[i] == 10)
			{
				digits[i] = digits[i] - 10;
				carry = 1;
			}
			else
				return digits;
		}
		digits[0] += carry;
		if(digits[0] == 10)
		{
			digits[0] -= 10;
			digits.insert(digits.begin(), 1);
		}
		return digits;
	}
};