leetcode

2.两数相加

python

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

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        carry = 0
        length = 0
        List_head = ListNode(0)
        List_head.next = l2
        List_card = List_head
        while(l1!=None and l2!=None):
            List_card = List_card.next
            
            sum_num = l1.val + l2.val + carry
            List_card.val = sum_num % 10
            carry = sum_num // 10
            
            l1 = l1.next
            l2 = l2.next
        
        while(l1!=None):
            List_card.next = ListNode(0)
            List_card = List_card.next
            
            sum_num = l1.val + carry
            List_card.val = sum_num % 10
            carry = sum_num // 10
            l1 = l1.next
        
        while(l2!=None):
            List_card = List_card.next
            
            sum_num = l2.val + carry
            List_card.val = sum_num % 10
            carry = sum_num // 10
            l2 = l2.next
        
        if carry>0:
            List_card.next = ListNode(0)
            List_card = List_card.next
            List_card.val = carry
            
        return List_head.next

写法更简单一些:

class Solution:    
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        prenode = ListNode(0)
        lastnode = prenode
        val = 0
        while val or l1 or l2:
            val, cur = divmod(val + (l1.val if l1 else 0) + (l2.val if l2 else 0), 10)
            lastnode.next = ListNode(cur)
            lastnode = lastnode.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        return prenode.next

其中 python divmod()
把除数和余数运算结果结合起来，返回一个包含商和余数的元组(a // b, a % b)

print(divmod(9, 2))
>> 4, 1

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* L_head = new ListNode(0);
        ListNode* List_card = L_head;
        int carry = 0;
        int sum;
        while((l1!=NULL)||(l2!=NULL)||(carry!=0)){
            sum = 0;
            
            if(l1!=NULL) {sum+=l1->val; l1=l1->next;}
            else l1=NULL;
            if(l2!=NULL) {sum+=l2->val; l2=l2->next;}
            else l2=NULL;
            sum+=carry;
            
            List_card->next = new ListNode(sum%10);
            List_card = List_card->next;
            carry = sum/10;
        }
        return L_head->next;
    }
};

9. 回文数

class Solution:
    def isPalindrome(self, x: int) -> bool:
        num_list=[]
        if(x<0): 
            return False
        while x is not 0:
            num_list.append(x%10)
            x = x//10
        
        n_list2 = num_list.copy()
        num_list.reverse()
        if n_list2 == num_list:
            return True
        else: return False

C++

class Solution {
public:
    bool isPalindrome(int x) {
        int reverseNum = 0;
        if(x<0||(x!=0&&x%10==0))
            return false;
        while(reverseNum<x){
            reverseNum = reverseNum*10 + x % 10;
            x = x / 10;
        }
        if(x == reverseNum ||x == reverseNum/10) return true;
        else return false;
    }
};

20.有效的括号

python

class Solution:
    def isValid(self, s: str) -> bool:
        left = ['(', '{', '[']
        right = [')', '}', ']']
        stack = []
        
        for i in s:
            if i in left:
                stack.append(i)
            if i in right:
                if len(stack)==0: return False
                else:
                    i_l = stack.pop()
                    if left.index(i_l)!=right.index(i): 
                        return False
        if len(stack)!=0:
            return False
        else:
            return True

C++

#include<stack>

class Solution {
public:
    bool isValid(string s) {
        stack<char> st;
        char r;
        for(int i=0; i<s.length(); i++){
           if(s[i]=='('||s[i]=='{'||s[i]=='[')
               st.push(s[i]);
           else
               if(st.empty()) return false;
               else{
                   r = st.top();
                   st.pop();
                   if(!((s[i]==')'&&r=='(')||(s[i]=='}'&&r=='{')||(s[i]==']'&&r=='['))) return false;
               }
        }
        if(!st.empty()) return false;
        else return true;
    }
};