【转载】面试题 02.05. 链表求和

题目链接

代码来自

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        ListNode p1 = l1, p2 = l2, result = new ListNode(0);
        ListNode p = result;
        int carr = 0;
        
        while(p1 != null || p2 != null || carr > 0)
        {
            int sum = carr;
            sum += p1 == null ? 0 : p1.val;
            sum += p2 == null ? 0 : p2.val;
            
            p.next = new ListNode(sum % 10);
            p = p.next;
            
            carr = sum / 10;
            
            if(p1 != null)
                p1 = p1.next;
            if(p2 != null)
                p2 = p2.next;
        }
        return result.next;
    }

改进代码

    public ListNode addTwoNumbers( ListNode l1 , ListNode l2 ){
        //目标链
        ListNode result = new ListNode( 0 );
        //指向目标链的可操作链（因为不能破坏原来的链）
        ListNode p = result;
        int sum = 0;
        //只要两条链中有一条链不为空，或者结果还有进位。就要一直循环
        while( l1 != null || l2 != null || sum > 0){
            //两个链的值，直接相加
            sum += l1 == null ? 0 : l1.val;
            sum += l2 == null ? 0 : l2.val;
            //取得末尾的数字，并且将其抛弃，形如  345 -> 34
            p.next = new ListNode( sum % 10 );
            p = p.next;
            
            sum = sum / 10;
            
            if( l1 != null ){
                l1 = l1.next;
            }
            if( l2 != null ){
                l2 = l2.next;
            }
        }
        return result.next;
    }