Plus One Linked List

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]

思路: 反转链表,然后+1,再反转链表;

/**
 * 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 null;
        }
        ListNode newhead = reverse(head);
        int carry = 1;
        ListNode node = newhead;
        while(carry != 0 ) {
            int sum = node.val + carry;
            carry = sum / 10;
            node.val = sum % 10;
            if(node.next != null) {
                node = node.next;
            } else {
                if(carry == 1) {
                    node.next = new ListNode(1);
                    break;
                }
            }
        }
        return reverse(newhead);
    }
    
    private ListNode reverse(ListNode node) {
        ListNode dummpy = new ListNode(0);
        dummpy.next = node;
        ListNode cur = node;
        while(cur!= null && cur.next != null) {
            ListNode nextnode = cur.next;
            cur.next = nextnode.next;
            nextnode.next = dummpy.next;
            dummpy.next = nextnode;
        }
        return dummpy.next;
    }
}

思路2:找到最后一个!=9的node,直接+1即可,如果没有,那么就证明全部是9,然后生成一个新node,后面的全部变为0;

/**
 * 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 null;
        }
        ListNode node = head;
        ListNode notnine = null;
        while(node != null) {
            if(node.val != 9) {
                notnine = node;
            }
            node = node.next;
        }
        
        if(notnine == null) {
            ListNode newhead = new ListNode(1);
            newhead.next = head;
            node = head;
            while(node != null){
                node.val = 0;
                node = node.next;
            }
            return newhead;
        } else {
            notnine.val += 1;
            node = notnine.next;
            while(node != null) {
                node.val = 0;
                node = node.next;
            }
            return head;
        }
    }
}

 

7-2 Mergin7-2 Merging Linked Lists 分数 15 作者 陈越 单位 浙江大学 Given two singly linked lists L 1 ​ =a 1 ​ →a 2 ​ →⋯→a n−1 ​ →a n ​ and L 2 ​ =b 1 ​ →b 2 ​ →⋯→b m−1 ​ →b m ​ . If n≥2m, you are supposed to reverse and merge the shorter one into the longer one to obtain a list like a 1 ​ →a 2 ​ →b m ​ →a 3 ​ →a 4 ​ →b m−1 ​ ⋯. For example, given one list being 6→7 and the other one 1→2→3→4→5, you must output 1→2→7→3→4→6→5. Input Specification: Each input file contains one test case. For each case, the first line contains the two addresses of the first nodes of L 1 ​ and L 2 ​ , plus a positive N (≤10 5 ) which is the total number of nodes given. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1. Then N lines follow, each describes a node in the format: Address Data Next where Address is the position of the node, Data is a positive integer no more than 10 5 , and Next is the position of the next node. It is guaranteed that no list is empty, and the longer list is at least twice as long as the shorter one. Output Specification: For each case, output in order the resulting linked list. Each node occupies a line, and is printed in the same format as in the input. Sample Input: 00100 01000 7 02233 2 34891 00100 6 00001 34891 3 10086 01000 1 02233 00033 5 -1 10086 4 00033 00001 7 -1 Sample Output: 01000 1 02233 02233 2 00001 00001 7 34891 34891 3 10086 10086 4 00100 00100 6 00033 00033 5 -1 代码长度限制 16 KB 时间限制 400 ms 内存限制 64 MB 栈限制g Linked Lists c语言
03-09
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