2. Add Two Numbers

最新推荐文章于 2021-12-02 14:05:46 发布

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本文介绍了一种解决链表数值相加的方法。该方法通过遍历两个链表，将链表节点上的数值相加并考虑进位，最终返回一个新的链表作为结果。文章提供了两种实现方式，一种使用额外的辅助节点进行链表构建，另一种则在一次循环中完成所有操作。

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

两个链表相加比较简单

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    if(l1 == null && l2 == null) return null;
    
    int carry = 0;
    ListNode pre = new ListNode(-1);
    ListNode dummy = pre;
    while(l1 != null || l2 != null){
        l1 = l1 == null ? new ListNode(0) : l1;
        l2 = l2 == null ? new ListNode(0) : l2;
        
        int sum = l1.val+l2.val+carry;
        int val = sum%10;
        carry = sum/10;
        ListNode cur = new ListNode(val);
        pre.next = cur;
        pre = cur;
        
        l1 = l1.next;
        l2 = l2.next;
    }
    
    if(carry > 0){
        ListNode cur = new ListNode(carry);
        pre.next = cur;
    }
    
    return dummy.next;
}

solution的更清晰

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(-1);
    ListNode cur = dummy;
    int carry = 0;
    while (l1 != null || l2 != null || carry != 0) {
        if (l1 != null) {
            carry += l1.val;
            l1 = l1.next;
        }
        if (l2 != null) {
            carry += l2.val;
            l2 = l2.next;
        }
        cur.next = new ListNode(carry % 10);
        carry /= 10;
        cur = cur.next;
    }
    return dummy.next;
}