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2.两数相加


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;
    }
} 
posted @ 2019-07-08 12:25  cznczai  阅读(140)  评论(0编辑  收藏  举报