leetcode 2816.翻倍以链表形式表示的数字
1.题目要求:
给你一个 非空 链表的头节点 head ，表示一个不含前导零的非负数整数。将链表 翻倍 后，返回头节点 head 。
2.题目代码:
/*** Definition for singly-linked list.* struct ListNode {* int val;* struct ListNode *next;* };*/
struct ListNode* doubleIt(struct ListNode* head){struct ListNode* cur = head;int count = 0;//1.遍历链表，得到结点个数while(cur){count++;cur = cur->next;}int* number = (int*)malloc(sizeof(int) * count);//2.根据结点个数用malloc申请数组int j = 0;cur = head;//3.把链表的结点放入数组while(cur){number[j] = cur->val;j++;cur = cur->next;}//4.给每个数组都乘以2for(int i = 0;i < j;i++){number[i] *= 2;}//5.进行进位for(int i = j - 1;i > 0;i--){if(number[i] > 9){number[i] %= 10;number[i - 1] += 1;}}//6.判断数组第一个数是否大于9if(number[0] > 9){struct ListNode* newnode = (struct ListNode*)malloc(sizeof(struct ListNode));newnode->val = number[0] / 10;newnode->next = head;head = newnode;cur = head;cur = cur->next;cur->val = number[0] % 10;int i = 1;cur = cur->next;while(cur){cur->val = number[i];i++;cur = cur->next;}return head;}else{cur = head;int i = 0;while(cur){cur->val = number[i];i++;cur = cur->next;}return head;}
}