2. Add Two Numbers【medium】
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
解法一：
1 class Solution { 2 public: 3 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { 4 ListNode * dummy = new ListNode(INT_MIN); 5 ListNode * head = dummy; 6 int carry = 0; 7 8 while (l1 != NULL && l2 != NULL) { 9 int sum = l1->val + l2->val + carry; 10 carry = sum / 10; 11 head->next = new ListNode(sum % 10); 12 head = head->next; 13 l1 = l1->next; 14 l2 = l2->next; 15 } 16 17 while (l1 != NULL) { 18 int sum = l1->val + carry; 19 carry = sum / 10; 20 head->next = new ListNode(sum % 10); 21 head = head->next; 22 l1 = l1->next; 23 } 24 25 while (l2 != NULL) { 26 int sum = l2->val + carry; 27 carry = sum / 10; 28 head->next = new ListNode(sum % 10); 29 head = head->next; 30 l2 = l2->next; 31 } 32 33 if (carry) { 34 head->next = new ListNode(carry); 35 head = head->next; 36 } 37 38 head->next = NULL; 39 40 return dummy->next; 41 } 42 };
写得太长了，下面有短码的方法
解法二：
1 public class Solution { 2 /** 3 * @param l1: the first list 4 * @param l2: the second list 5 * @return: the sum list of l1 and l2 6 */ 7 public ListNode addLists(ListNode l1, ListNode l2) { 8 // write your code here 9 ListNode dummy = new ListNode(0); 10 ListNode tail = dummy; 11 12 int carry = 0; 13 for (ListNode i = l1, j = l2; i != null || j != null; ) { 14 int sum = carry; 15 sum += (i != null) ? i.val : 0; 16 sum += (j != null) ? j.val : 0; 17 18 tail.next = new ListNode(sum % 10); 19 tail = tail.next; 20 21 carry = sum / 10; 22 i = (i == null) ? i : i.next; 23 j = (j == null) ? j : j.next; 24 } 25 26 if (carry != 0) { 27 tail.next = new ListNode(carry); 28 } 29 return dummy.next; 30 } 31 }
参考了九章的代码
解法三：
1 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { 2 ListNode preHead(0), *p = &preHead; 3 int extra = 0; 4 while (l1 || l2 || extra) { 5 int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra; 6 extra = sum / 10; 7 p->next = new ListNode(sum % 10); 8 p = p->next; 9 l1 = l1 ? l1->next : l1; 10 l2 = l2 ? l2->next : l2; 11 } 12 return preHead.next; 13 }
参考了@ce2 的代码
解法四：
1 class Solution { 2 public: 3 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { 4 // 题意可以认为是实现高精度加法 5 ListNode *head = new ListNode(0); 6 ListNode *ptr = head; 7 int carry = 0; 8 while (true) { 9 if (l1 != NULL) { 10 carry += l1->val; 11 l1 = l1->next; 12 } 13 if (l2 != NULL) { 14 carry += l2->val; 15 l2 = l2->next; 16 } 17 ptr->val = carry % 10; 18 carry /= 10; 19 // 当两个表非空或者仍有进位时需要继续运算，否则退出循环 20 if (l1 != NULL || l2 != NULL || carry != 0) { 21 ptr = (ptr->next = new ListNode(0)); 22 } else break; 23 } 24 return head; 25 } 26 };
参考了九章的代码