LeetCode 2. Add Two Numbers
原题链接：https://leetcode.com/problems/add-two-numbers/
题目：
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
题解：
思路两个val和进位carry相加，组成新点往后连，注意两个list长度不同和最后是否还有一个进位的情况。
Time Complexity: O(n), n是较长list的长度.
Space: O(n).
AC Java:
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution { 10 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 11 if(l1 == null){ 12 return l2; 13 } 14 if(l2 == null){ 15 return l1; 16 } 17 18 ListNode dummy = new ListNode(0); 19 ListNode cur = dummy; 20 int carry = 0; 21 22 while(l1 != null || l2 != null){ 23 if(l1 != null){ 24 carry += l1.val; 25 l1 = l1.next; 26 } 27 if(l2 != null){ 28 carry += l2.val; 29 l2 = l2.next; 30 } 31 32 cur.next = new ListNode(carry%10); 33 cur = cur.next; 34 carry = carry/10; 35 } 36 if(carry != 0){ 37 cur.next = new ListNode(carry); 38 } 39 return dummy.next; 40 } 41 }
AC C++:
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode() : val(0), next(nullptr) {} 7 * ListNode(int x) : val(x), next(nullptr) {} 8 * ListNode(int x, ListNode *next) : val(x), next(next) {} 9 * }; 10 */ 11 class Solution { 12 public: 13 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { 14 if(!l1){ 15 return l2; 16 } 17 18 if(!l2){ 19 return l1; 20 } 21 22 ListNode* dummy = new ListNode(-1); 23 ListNode* cur = dummy; 24 int carry = 0; 25 while(l1 || l2){ 26 if(l1){ 27 carry += l1->val; 28 l1 = l1->next; 29 } 30 31 if(l2){ 32 carry += l2->val; 33 l2 = l2->next; 34 } 35 36 cur->next = new ListNode(carry % 10); 37 cur = cur->next; 38 carry /= 10; 39 } 40 41 if(carry != 0){ 42 cur->next = new ListNode(carry); 43 } 44 45 return dummy->next; 46 } 47 };
也可以选择更改较长list的node val.
不需要n的extra space.
Time Complexity: O(n).
Space: O(1).
AC Java:
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution { 10 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 11 if(l1 == null){ 12 return l2; 13 } 14 if(l2 == null){ 15 return l1; 16 } 17 int len1 = 0; 18 ListNode cur = l1; 19 while(cur != null){ 20 cur = cur.next; 21 len1++; 22 } 23 24 int len2 = 0; 25 cur = l2; 26 while(cur != null){ 27 cur = cur.next; 28 len2++; 29 } 30 31 ListNode dummy = new ListNode(0); 32 if(len1 > len2){ 33 dummy.next = l1; 34 }else{ 35 dummy.next = l2; 36 } 37 cur = dummy; 38 int carry = 0; 39 40 while(l1 != null || l2 != null){ 41 if(l1 != null){ 42 carry += l1.val; 43 l1 = l1.next; 44 } 45 if(l2 != null){ 46 carry += l2.val; 47 l2 = l2.next; 48 } 49 cur.next.val = carry%10; 50 cur = cur.next; 51 carry /= 10; 52 } 53 if(carry != 0){ 54 cur.next = new ListNode(1); 55 } 56 return dummy.next; 57 } 58 }
