Given a non-negative integer represented as non-empty a singly linked list of digits, plus one to the integer.
You may assume the integer do not contain any leading zero, except the number 0 itself.
The digits are stored such that the most significant digit is at the head of the list.
Example:
Input:
1->2->3
Output:
1->2->4
一刷
题解:
用递归来寻找最底层是否要进位
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode plusOne(ListNode head) {
if(head == null) return head;
if(exceed(head.next) || head.next == null){
if(head.val == 9){
ListNode newHead = new ListNode(1);
newHead.next = head;
head.val = 0;
return newHead;
}else{
head.val++;
return head;
}
}else{
return head;
}
}
private boolean exceed(ListNode node){
if(node == null) return false;
if(node.next == null || exceed(node.next)){
if(node.val == 9){
node.val = 0;
return true;
}else{
node.val++;
return false;
}
}
return false;
}
}
二刷
简化写法:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode plusOne(ListNode head) {
ListNode Dummy = new ListNode(0);
Dummy.next = head;
if(helper(head)){
Dummy.val++;
return Dummy;
}else{
return head;
}
}
private boolean helper(ListNode node){
if(node == null) return true;
if(helper(node.next)){
if(node.val == 9 ){
node.val = 0;
return true;
}
node.val++;
}
return false;
}
}
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