LeetCode链表练习
题目来自于:https://leetcode-cn.com/problems/add-two-numbers
Python
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: ls = ListNode() ls_copy = ls carry = 0 while l1 or l2: if not l1: sum_bit = l2.val + carry l2 = l2.next elif not l2: sum_bit = l1.val + carry l1 = l1.next else: sum_bit = l2.val + l1.val + carry l1 = l1.next l2 = l2.next if sum_bit >= 10: ls_copy.next = ListNode(sum_bit-10) carry = 1 else: ls_copy.next = ListNode(sum_bit) carry = 0 ls_copy = ls_copy.next if carry == 1: ls_copy.next = ListNode(1) return ls.next
C语言
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){ struct ListNode * ls=(struct ListNode*)malloc(sizeof(struct ListNode));//创建一个头结点 struct ListNode * ls_copy=ls;//声明一个指针指向头结点,用于遍历链表 int carry = 0; int sum_bit; while (l1 || l2){ if (l1==NULL){ sum_bit = l2->val + carry; l2 = l2->next; }else if (l2==NULL){ sum_bit = l1->val + carry; l1 = l1->next; }else{ sum_bit = l1->val + l2->val + carry; l1 = l1->next; l2 = l2->next; } struct ListNode * c=(struct ListNode*)malloc(sizeof(struct ListNode)); if (sum_bit >= 10){ c->val = sum_bit-10; carry = 1; }else{ c->val = sum_bit; carry = 0; } c->next = NULL; ls_copy->next = c; ls_copy = ls_copy->next; } if (carry == 1){ struct ListNode * c=(struct ListNode*)malloc(sizeof(struct ListNode)); c->val = 1; c->next = NULL; ls_copy->next = c; } return ls->next; }
思路
参考于评论区的大佬。
1.首先创建一个空的链表,用于存储计算的结果。
2.如果待计算的l1和l2中有一个未到末尾,则执行计算。
2.1 如果l1和l2都未到末尾,则直接将当前值相加,若其中一个已到达末尾,则不加其数值。
2.2 将l1和l2向后移动一位,如果已到末尾则不移动。
2.3 根据求和值是否大于等于10,酌情进位。
3.存储求和结果。
4.如果循环结束时进位为1,则在结果最后再补一位1.