2.两数相加

标签：ima   public   add   ddt   number   com   就是   node   init   

下面是我的算法
一开始采用的是while(ln2 != null && ln1 != null) 结果答案就是解不出
然后采用下面更加啰嗦的 就好了 ~~~~

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode ln = head;
        ListNode ln1 = l1;
        ListNode ln2 = l2;
        int temp = 0;
        while (ln1 != null || ln2 != null) {
            if (ln1 == null) {
                ln.next = new ListNode((ln2.val + temp) % 10);
                temp = (ln2.val + temp) / 10;
                ln = ln.next;
            }
            else if (ln2 == null) {
                ln.next = new ListNode((ln1.val + temp) % 10);
                temp = (ln1.val + temp) / 10;
                ln = ln.next;
            } 
            else {
                ln.next = new ListNode((ln1.val + ln2.val + temp) % 10);
                temp = (ln1.val + ln2.val + temp) / 10;
                ln = ln.next;
            }
        }
        if (temp != 0)
            ln.next = new ListNode(1);
        return head.next;
    }
}

简化版：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode ln1 = l1, ln2 = l2, head = null, node = null;
        int carry = 0, remainder = 0, sum = 0;
        head = node = new ListNode(0);
        
        while(ln1 != null || ln2 != null || carry != 0) {
            sum = (ln1 != null ? ln1.val : 0) + (ln2 != null ? ln2.val : 0) + carry;
            carry = sum / 10;
            remainder = sum % 10;
            node = node.next = new ListNode(remainder);
            ln1 = (ln1 != null ? ln1.next : null);
            ln2 = (ln2 != null ? ln2.next : null);
        }
        return head.next;
    }
} 

2.两数相加

标签：ima   public   add   ddt   number   com   就是   node   init   

原文地址：https://www.cnblogs.com/cznczai/p/11150426.html