【LeetCode】Add Two Numbers

问题描述

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

译：给定两个用列表表示的非负整数，整数逆序存储并且每个节点存储一个整数位，求两个数的和并用列表返回。

算法思想

整数加法问题，需要考虑进位和余数问题，
利用变量carryBit = 0存储进位。
1. 如果l1!=null && l2!=null，那么计算（节点及进位的和%10）添加到r中并计算新的进位
2. 如果l1!=null,那么把l1的数加进位添加到r中
3. 如果l2!=null,那么把l2的数加进位添加到r中
4. 如果l1==null&&l2==null&&carryBit!=null，把1添加到r中

算法实现

/**
 * Definition for singly-linked list.
 */
public class Solution {
    public class ListNode {
        int val;
        ListNode next;
        ListNode(int x) {
            val = x;
        }
    }
    public static ListNode tmpL1 = null;
    public static ListNode tmpL2 = null;
    public static ListNode sumListNode = null;

    public Solution(){
        tmpL1 = new ListNode(2);
        tmpL1.next = new ListNode(4);
        tmpL1.next.next = new ListNode(8);
        tmpL1.next.next.next = new ListNode(9);

        tmpL2 = new ListNode(5);
        tmpL2.next = new ListNode(6);
        tmpL2.next.next = new ListNode(4);
    }
    /**
     * carryBit = 0;
     * 如果l1!=null && l2!=null，那么计算和+进位添加到r中并计算新的进位
     * 如果l1!=null,那么把l1的数加进位添加到r中
     * 如果l2!=null,那么把l2的数加进位添加到r中
     * 如果l1==null&&l2==null&&carryBit!=null，把1添加到r中
     * 
     * @param l1
     * @param l2
     * @return
     */
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int carryBit = 0,tmpBit = 0,tmpSum = 0;
        ListNode tmpL1=l1,tmpL2=l2,sumListHead = null,sumListNode = new ListNode(0);
        sumListHead = sumListNode;
        while(tmpL1!=null&&tmpL2!=null){
            tmpSum = tmpL1.val+tmpL2.val+carryBit;
            tmpBit = tmpSum%10;
            carryBit = tmpSum/10;
            ListNode newBit = new ListNode(tmpBit);
            sumListNode.next = newBit;
            sumListNode = sumListNode.next;
            tmpL1 = tmpL1.next;
            tmpL2 = tmpL2.next;
        }

        ListNode tmpL = null;
        if(tmpL1!=null) tmpL = tmpL1;
        if(tmpL2!=null) tmpL = tmpL2;

        while(tmpL!=null){
            tmpSum = tmpL.val+carryBit;
            tmpBit = tmpSum%10;
            carryBit = tmpSum/10;
            ListNode newBit = new ListNode(tmpBit);
            sumListNode.next = newBit;
            sumListNode = sumListNode.next;
            tmpL = tmpL.next;
        }

        if(carryBit == 1){
            ListNode newBit = new ListNode(1);
            sumListNode.next = newBit;
            sumListNode = sumListNode.next;
        }

        return sumListHead.next;
    }
    public static void main(String [] args){
        //Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
        //Output: 7 -> 0 -> 8
        Solution s = new Solution();
        sumListNode = s.addTwoNumbers(tmpL1,tmpL2);
        while(sumListNode!=null){
            System.out.println(sumListNode.val);
            sumListNode = sumListNode.next;
        }

    }
}

算法时间

f(n) = O(L1.length+L2.length)

演示结果

7
0
3
0
1