Add Two Numbers I & II
Add Two Numbers I
You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.
Given 7->1->6 + 5->9->2. That is, 617 + 295.
Return 2->1->9. That is 912.
Given 3->1->5 and 5->9->2, return 8->0->8.
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) return l2; 12 if (l2 == null) return l1; 13 14 int value1 = 0, value2 = 0, carry = 0, sum = 0; 15 ListNode head = new ListNode(0); 16 ListNode current = head; 17 18 while (l1 != null || l2 != null || carry == 1) { 19 value1 = (l1 == null ? 0 : l1.val); 20 value2 = (l2 == null ? 0 : l2.val); 21 sum = value1 + value2 + carry; 22 23 current.next = new ListNode(sum % 10); 24 current = current.next; 25 26 carry = sum / 10; 27 l1 = (l1 == null) ? null : l1.next; 28 l2 = (l2 == null) ? null : l2.next; 29 } 30 return head.next; 31 } 32 }
Add Two Numbers II
You are given two linked lists representing two non-negative numbers. The most significant digit comes first 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.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:
Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 8 -> 0 -> 7
分析: reverse 之后再相加，然后再reverse result linkedlist
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) return l2; 12 if (l2 == null) return l1; 13 14 l1 = reverse(l1); 15 l2 = reverse(l2); 16 17 return reverse(helper(l1, l2)); 18 } 19 20 public ListNode reverse(ListNode head) { 21 if (head == null || head.next == null) return head; 22 23 ListNode pre = null; 24 ListNode current = head; 25 ListNode next = null; 26 27 while (current != null) { 28 next = current.next; 29 current.next = pre; 30 pre = current; 31 current = next; 32 } 33 return pre; 34 } 35 36 public ListNode helper(ListNode l1, ListNode l2) { 37 if (l1 == null) return l2; 38 if (l2 == null) return l1; 39 40 int value1 = 0, value2 = 0, carry = 0, sum = 0; 41 ListNode head = new ListNode(0); 42 ListNode current = head; 43 44 while (l1 != null || l2 != null || carry == 1) { 45 value1 = (l1 == null ? 0 : l1.val); 46 value2 = (l2 == null ? 0 : l2.val); 47 sum = value1 + value2 + carry; 48 49 current.next = new ListNode(sum % 10); 50 current = current.next; 51 52 carry = sum / 10; 53 l1 = (l1 == null) ? null : l1.next; 54 l2 = (l2 == null) ? null : l2.next; 55 } 56 return head.next; 57 } 59 }
方法二：用stack存list中的值，这样从上到下就是位数从小到大。
1 public class Solution { 2 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 3 Stack<Integer> s1 = new Stack<>(); 4 Stack<Integer> s2 = new Stack<>(); 5 6 while (l1 != null) { 7 s1.push(l1.val); 8 l1 = l1.next; 9 } 10 11 while (l2 != null) { 12 s2.push(l2.val); 13 l2 = l2.next; 14 } 15 int carry = 0; 16 ListNode head = new ListNode(0); 17 while (!s1.empty() || !s2.empty() || carry != 0) { 18 int sum = 0; 19 if (!s1.empty()) { 20 sum += s1.pop(); 21 } 22 if (!s2.empty()) { 23 sum += s2.pop(); 24 } 25 sum += carry; 26 carry = sum / 10; 27 ListNode node = new ListNode(sum % 10); 28 node.next = head.next; 29 head.next = node; 30 } 31 return head.next; 32 } 33 }
