2

思路

这道题我大概看了下, 没什么特别的感觉啊, 首先既然list中的digit是倒序存放的, 也就是从头部开始一次表示个位, 十位, 百位... 这样的话就简单了, 直接按照加过去, 只需要O(n)的时间.

实现

具体实现上, 也没什么, 就是不断地对于list进行循环然后直至到底list末尾, 要注意的是list可能不等长, 还有如果在两个list都到达末尾的情况下累加器里面还有值(上一次计算有进位), 那么还需要再添加一个node.

提交

这次是一次AC

代码

里面有一些我自己写的测试代码.

/**
 * Definition for singly-linked list.
 */
public class Solution {
    public class ListNode {
        int val;
        ListNode next;
        ListNode(int x) { val = x; }
    }

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode re = null, cur = null;
        int adder = 0;
        while(l1 != null && l2 != null){
            int sum = l1.val + l2.val + adder;
            int digit = sum % 10;
            adder = sum / 10;
            if(re == null){
                re = new ListNode(digit);
                cur = re;
            }
            else{
                cur.next = new ListNode(digit);
                cur = cur.next;
            }
            l1 = l1.next;
            l2 = l2.next;
        }

        ListNode l3 = l1 == null ? l2 : l1;
        while(l3 != null){
            int sum = l3.val + adder;
            int digit = sum % 10;
            adder = sum / 10;
            if(re == null){
                re = new ListNode(digit);
                cur = re;
            }
            else{
                cur.next = new ListNode(digit);
                cur = cur.next;
            }
            l3 = l3.next;
        }

        if(adder != 0){
            cur.next = new ListNode(adder);
        }

        return re;
    }

    private ListNode getNode(int[] list){
        if(list.length == 0) return null;
        ListNode re = new ListNode(list[0]);
        ListNode cur = re;
        for(int i = 1; i < list.length; ++i){
            cur.next = new ListNode(list[i]);
            cur = cur.next;
        }
        return re;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        ListNode l1 = solution.getNode(new int[]{2, 4, 3});
        ListNode l2 = solution.getNode(new int[]{5, 6, 0});
        ListNode l3 = solution.addTwoNumbers(l1, l2);

        while(l3 !=null){
            System.out.println(l3.val);
            l3 = l3.next;
        }
    }
}

最佳实现

这道题基本上没什么优化了, 但是代码上还可以写得更整洁, 第一个while其实也能写到第二个while里面, 这里有一份样例代码 :

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;
}