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  • LeetCode 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

    题意:将两个用链表逆序存储的数进行加法运算,并将结果也用链表逆序存储。如上例的342 + 564 = 807.

    题解:掌握链表的头插法和尾插法,从左往右每个节点进行加法运行,进位带入下个个节点运算。用尾插法保存结果数据。要注意节点的空判断,以防出现空指针,当计算最后的进位不为0时,要再加入一个节点存放进位。

    复制代码
    class ListNode {
          int val;
          ListNode next;
          ListNode(int x) { val = x; }
     }
    
    public class AddTwoNumbers {
        
        public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
            ListNode res = new ListNode(0);
            ListNode r = res;
            int x = l1.val;
            int y = l2.val;
            res.val = ( x + y ) % 10;
            int ten = ( x + y ) / 10;
            l1 = l1.next;
            l2 = l2.next;
            
            while(l1 != null || l2 !=null ){
                // 防止出现空指针
                if(l1 == null){
                    l1 = new ListNode(0);
                    x = 0;
                }else{
                    x = l1.val;
                }
                
                if(l2 == null){
                    l2 = new ListNode(0);
                    y = 0;
                }else{
                    y = l2.val;
                }
                
                int single = ( x + y + ten ) % 10;
                ListNode tmp = new ListNode(single);
                r.next = tmp;
                r = tmp;
                ten = ( x + y + ten ) / 10;
                l1 = l1.next;
                l2 = l2.next;
            }
            //存放最后的进位
            if(ten != 0){
                r.next = new ListNode(ten);
            }
            return res;
        }
        
        public static void main(String[] args) {
            // TODO Auto-generated method stub
            int a[] = {5};
            int b[] = {5};
            
            ListNode la = new ListNode(0);
            la.val = a[0];
            for(int i = a.length -1 ;i>0;i--){
                ListNode next = new ListNode(a[i]);    
                next.next = la.next;
                la.next = next;
            }
            
            ListNode lb = new ListNode(0);
            lb.val = b[0];
            
            for(int i = b.length -1 ;i > 0 ;i--){
                ListNode next = new ListNode(b[i]);    
                next.next = lb.next;
                lb.next = next;
            }
            
            ListNode c = addTwoNumbers(la,lb);
            
            while( c != null){
                System.out.print(c.val);
                if(c.next != null){
                    System.out.print("-->");
                }
                c = c.next;
            }
        }
    }
    复制代码
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  • 原文地址:https://www.cnblogs.com/AndyDai/p/4810864.html
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