题目445. 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
思路:类似用头插法建立链表
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if(l1 == null || l2 == null){
return l1 == null ? l1 : l2;
}
Stack<ListNode> stack1 = new Stack<ListNode>();
Stack<ListNode> stack2 = new Stack<ListNode>();
saveList(stack1,l1);
saveList(stack2,l2);
ListNode tempHead = new ListNode(1);
ListNode node1 = null;
ListNode node2 = null;
int rise = 0;
while(!stack1.empty() && !stack2.empty()){
node1 = stack1.pop();
node2 = stack2.pop();
int sum = node1.val + node2.val + rise;
rise = sum / 10;
sum %= 10;
ListNode newNode = new ListNode(sum);
newNode.next = tempHead.next;
tempHead.next = newNode;
}
while(!stack1.empty()){
node1 = stack1.pop();
int sum = node1.val + rise;
rise = sum / 10;
sum %= 10;
ListNode newNode = new ListNode(sum);
newNode.next = tempHead.next;
tempHead.next = newNode;
}
while(!stack2.empty()){
node2 = stack2.pop();
int sum = node2.val + rise;
rise = sum / 10;
sum %= 10;
ListNode newNode = new ListNode(sum);
newNode.next = tempHead.next;
tempHead.next = newNode;
}
if(rise != 0){
ListNode newNode = new ListNode(rise);
newNode.next = tempHead.next;
tempHead.next = newNode;
}
return tempHead.next;
}
private void saveList(Stack<ListNode> stack, ListNode head){
ListNode node = head;
while(node != null){
stack.push(node);
node = node.next;
}
}
}
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