leetcode-002 Add Two Numbers

• P002-Add-Two-Numbers
  • 思路分析
  • 代码
    • java
    • python

P002-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: ( ->  -> ) + ( ->  -> )
Output:  ->  -> 
           

思路分析

就是两个链表的并行遍历，并不怎么难。

• 肯定得遍历两个链表
• 两链表长度相同时，对应位置相加并将进位保留到下一次循环
• 两链表长度不相同时，往长度较短的链表尾部补零(最高位和现实生活中的数字恰好相反)

代码

java

/***
 * 2 4 3        2 4 3 0     1 2 3 4 0
 * 5 6 4        5 6 4 9     6 7 8 9 9
 * -------      -------     ----------
 * 7 0 8        7 0 8 9     7 9 1 4 0 1
 * 
 * **/
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    if (l1 == null)
        return l2;
    if (l2 == null)
        return l1;
    ListNode l = l1;
    ListNode r = l2;
    ListNode ret = new ListNode();
    int up = ;
    int currentVal = ;
    ListNode last = ret;
    while (l != null || r != null) {

        // 两个链表长度不一致：补零
        if (l == null)
            l = new ListNode();
        if (r == null)
            r = new ListNode();

        // 当前值:加上进位
        currentVal = (l.val + r.val) + up;

        if (currentVal >= ) {
            up = currentVal / ;// 为下一轮提供的进位
            currentVal = currentVal % ;// 当前值
        } else {
            up = ;
        }

        // 新节点
        ListNode newNode = new ListNode(currentVal);

        last.next = newNode;
        last = last.next;

        l = l.next;
        r = r.next;
    }
    if (up != ) {
        last.next = new ListNode(up);
    }
    return ret.next;
}
           

python

def addTwoNumbers(self, l1, l2):
    """
    :type l1: ListNode
    :type l2: ListNode
    :rtype: ListNode
    """
    if (not l1) : return l2
    if (not l2) : return l1
    l = l1;r = l2;
    ret = ListNode();last = ret
    up = ;currentVal = 

    while (l or r):
        if not l:l = ListNode()
        if not r:r = ListNode()

        currentVal = (l.val + r.val) + up

        if currentVal >= :
            up = currentVal / 
            currentVal = currentVal % 
        else:
            up = 

        newNode = ListNode(currentVal)
        last.next = newNode
        last = last.next

        l = l.next
        r = r.next

    # end while

    if up != :
        last.next = ListNode(up)

    return ret.next