Add Two Numbers 把两个以链表形式表示的数相加 @LeetCode

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本文介绍了一种解决两数相加问题的方法，通过链表形式存储数字，并以逆序方式表示数值。该方法使用三个指针进行操作，能够处理不同长度的链表，并考虑了进位情况。

思路：

新建结果链表，用3个指针处理，注意carry

package Level3;

import Utility.ListNode;

/**
 * 
 * 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
 */
public class S2 {

	public static void main(String[] args) {
		ListNode l1 = new ListNode(2);
		ListNode x2 = new ListNode(4);
		ListNode x3 = new ListNode(3);
		l1.next = x2;
		x2.next = x3;
		
		ListNode l2 = new ListNode(5);
		ListNode y2 = new ListNode(6);
		ListNode y3 = new ListNode(4);
		l2.next = y2;
		y2.next = y3;
		
		ListNode ret = addTwoNumbers(l1, l2);
		ret.print();
	}
	
	public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		int carry = 0;			// 进位
		ListNode p1 = l1;		// 标记l1的位置
		ListNode p2 = l2;		// 标记l2的位置
		ListNode ret = null;	// 返回链表
		ListNode p3 = ret;	// 标记返回链表的位置
		
		// 处理相同长度部分
		while(p1!=null && p2!=null){
			int add = p1.val + p2.val + carry;
			if(add < 10){
				carry = 0;
			}else{
				add -= 10;
				carry = 1;
			}
			ListNode n = new ListNode(add);
			if(ret == null){
				ret = n;
				p3 = n;
			}else{
				p3.next = n;
				p3 = n;
			}
			p1 = p1.next;
			p2 = p2.next;
		}

		// 处理不同长度部分
		while(p1 != null){
			int add = p1.val + carry;
			if(add < 10){
				carry = 0;
			}else{
				add -= 10;
				carry = 1;
			}
			ListNode n = new ListNode(add);
			p3.next = n;
			p3 = n;
			p1 = p1.next;
		}
		
		while(p2 != null){
			int add = p2.val + carry;
			if(add < 10){
				carry = 0;
			}else{
				add -= 10;
				carry = 1;
			}
			ListNode n = new ListNode(add);
			p3.next = n;
			p3 = n;
			p2 = p2.next;
		}
		
		// 处理最后可能会出现的进位
		if(p1==null && p2==null && carry!=0){
			ListNode n = new ListNode(carry);
			p3.next = n;
		}
		
		return ret;
	}

}

同合并链表题

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode cur1 = l1, cur2 = l2;
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        int carry = 0;
        while(cur1 != null && cur2 != null) {
            int sum = cur1.val + cur2.val + carry;
            if(sum >= 10) {
                carry = 1;
                cur.next = new ListNode(sum-10);
            } else {
                carry = 0;
                cur.next = new ListNode(sum);
            }
            cur1 = cur1.next;
            cur2 = cur2.next;
            cur = cur.next;
        }
        
        while(cur1 != null) {
            int sum = cur1.val + carry;
            if(sum >= 10) {
                carry = 1;
                cur.next = new ListNode(sum-10);
            } else {
                carry = 0;
                cur.next = new ListNode(sum);
            }
            cur1 = cur1.next;
            cur = cur.next;
        }
        
        while(cur2 != null) {
            int sum = cur2.val + carry;
            if(sum >= 10) {
                carry = 1;
                cur.next = new ListNode(sum-10);
            } else {
                carry = 0;
                cur.next = new ListNode(sum);
            }
            cur2 = cur2.next;
            cur = cur.next;
        }
        
        if(carry == 1) {
            cur.next = new ListNode(1);
        }
        
        return dummy.next;
    }
}