leetCode 2 Add Two Numbers

问题：Add Two Numbers

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

解题过程： 在解题的过程中首先因为是数的反序，故先将链表反转，然后相加，同时比较链表长度，如不一致，则需表头补0。

不过后面提交的过程中，发现完全不需要反转链表，主要是243+564->708,狗日的342+465->708是相同的，造成要反转。

解题代码：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {       //ListNode pre1 = l1,cur = l1, next = null,pre2 = l2;       //int i = 0,j = 0;       //艹，发现想多了，返回数据根本不需要反转              //l1翻转//        while(cur != null){//            next = cur;//            cur = cur.next;//            next.next = pre1;//            pre1 = next;//            i++;//记录l1位数//        }//        l1.next = null;//        //        //l2反转//        cur = l2;//        while(cur != null){//            next = cur;//            cur = cur.next;//            next.next = pre2;//            pre2 = next;//            j++;//记录l2位数//        }//        l2.next = null;       //System.out.println("i:" + i);       //System.out.println("j:" + j);       //如果l1大于l2位数，则l2头结点补0；反之亦然//        if(i > j){//         for(int k = 0; k < i-j; k++){//         ListNode ln = new ListNode(0); //         ln.next = pre2;//         pre2 = ln;//        }//        }//        else{//         for(int k = 0; k < j-i; k++){//         ListNode ln = new ListNode(0); //         ln.next = pre1;//         pre1 = ln;//        }//        }              //两数相加       int addNum = 0;       ListNode last = null;       ListNode head = null;       boolean isOver = false;//是否进位       while(l1 != null && l2 != null){        if(isOver){        addNum = l1.val + l2.val + 1;        isOver = false;        }        else{        addNum = l1.val + l2.val;        }        if(addNum > 9){        isOver = true;        addNum = addNum % 10;//取个位数        }        ListNode ln = new ListNode(addNum);         if(head == null){        last = head = ln;        }else{        last.next = ln;        last = ln;        }        l1 = l1.next;        l2 = l2.next;       }              //如果l1,l2位数不相同，则必有且只有1个还有数据       while(l1 != null){        //如果有进位        if(isOver){        l1.val += 1;        isOver = false;        if(l1.val > 9){        isOver = true;        l1.val = l1.val % 10;//取个位数        }        }        last.next = l1;        last = l1;        l1 = l1.next;       }       while(l2 != null){        //如果有进位        if(isOver){        l2.val += 1;        isOver = false;        if(l2.val > 9){        isOver = true;        l2.val = l2.val % 10;//取个位数        }        }        last.next = l2;        last = l2;        l2 = l2.next;       }       //如果还有进位未处理       if(isOver){        ListNode ln = new ListNode(1);         last.next = ln;        last = ln;       }              //如果位数不相等，则把剩余的相加最后      // while(head != null)      // {     //  System.out.println(head.val);     //  head = head.next;      // }return head;   }}

上面的代码用了递归排序的思想，不过实现太罗嗦了，下面为简洁的实现方式：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {       //两数相加       int addNum = 0;       ListNode last = new ListNode(0);       ListNode head = last;       int isOver = 0;//是否进位       while(true){        int v1 = l1 == null ? 0 : l1.val;        int v2 = l2 == null ? 0 : l2.val;                addNum = v1 + v2 + isOver;        isOver = addNum / 10;        addNum = addNum % 10;                last.val = addNum;                if(l1 != null)        <span style="white-space:pre"></span>l1 = l1.next;        if(l2 != null)        <span style="white-space:pre"></span>l2 = l2.next;                if(l1 == null && l2 == null && isOver == 0){        <span style="white-space:pre"></span>break;        }        else{        <span style="white-space:pre"></span>last.next = new ListNode(0);        <span style="white-space:pre"></span>last = last.next;        }       }return head;   }}