LeetCode - 025. 链表中的两数相加

方法一：通过链表反转 + 分情况

首先，我们将链表分为一长一短，由此链表相加可以分为三种情况：

我们用curS表示当前短链表走到的位置，curL表示当前长链表走到的位置，carray表示进位，last用于存储null的上一个节点，方便当长短链表都走完而carry不为0时，新增节点

①长链表没走完，短链表也没走完

//有长有短
while(curS != null){
    
    int sum = curL.val + curS.val + carray;
    curL.val = sum % 10;
    carray = sum / 10;
    last = curL;//存储null的上一个节点
    curL = curL.next;
    curS = curS.next;
}

②只有长

//有长无短
while(curL != null){
    
    int sum = curL.val + carray;
    curL.val = sum % 10;
    carray = sum / 10;
    last = curL;
    curL = curL.next;
}

③无长无短

//无长无短
if(carray != 0){
    
    last.next = new ListNode(1);
}

④反转链表（注意最后结果也需要反转）

public ListNode reverse(ListNode head){
    
    if(head == null){
    
        return null;
    }
    ListNode pre = null;
    ListNode cur = head;
    ListNode next = cur.next;
    while(cur != null){
    
        next = cur.next;
        cur.next = pre;
        pre = cur;
        cur = next;
    }
    return pre;
}

全部代码：

class Solution {
    
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    
        //获取两链表长度
        int len1 = getLength(l1);
        int len2 = getLength(l2);
        l1 = reverse(l1);
        l2 = reverse(l2);
        ListNode l = len1 >= len2 ? l1 : l2;//长链表
        ListNode s = len1 < len2 ? l1 : l2;
        ListNode curL = l;
        ListNode curS = s;
        ListNode last = curL;
        int carray = 0;//进位
        int sum = 0;
        //有长有短
        while(curS != null){
    
            sum = curL.val + curS.val + carray;
            curL.val = sum % 10;
            carray = sum / 10;
            last = curL;//存储null的上一个节点
            curL = curL.next;
            curS = curS.next;
        }
        //有长无短
        while(curL != null){
    
            sum = curL.val + carray;
            curL.val = sum % 10;
            carray = sum / 10;
            last = curL;
            curL = curL.next;
        }
        //无长无短
        if(carray != 0){
    
            last.next = new ListNode(1);
        }
        return reverse(l);
    }
    public int getLength(ListNode l){
    
        int res = 0;
        if(l == null) return res;
        while(l != null){
    
            res++;
            l = l.next;
        }
        return res;
    }
    public ListNode reverse(ListNode head){
    
        if(head == null){
    
            return null;
        }
        ListNode pre = null;
        ListNode cur = head;
        ListNode next = cur.next;
        while(cur != null){
    
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}

方法二：通过Deque实现【使用栈的特性实现】

首先，将两个不同的链表分别入不同的栈，再利用栈的特性分别出栈

class Solution {
    
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    
        Deque<Integer> stack1 = new ArrayDeque<>();
        Deque<Integer> stack2 = new ArrayDeque<>(); 
        while(l1 != null){
    
            stack1.push(l1.val);
            l1 = l1.next;
        }
        while(l2 != null){
    
            stack2.push(l2.val);
            l2 = l2.next;
        }
        int carray = 0;//进位
        ListNode res = null;
        while(!stack1.isEmpty() || !stack2.isEmpty() || carray != 0){
    
            int a = stack1.isEmpty() ? 0 : stack1.pop();
            int b = stack2.isEmpty() ? 0 : stack2.pop();
            int sum = a + b + carray;
            carray = sum / 10;
            int val = sum % 10;
            ListNode cur = new ListNode(val);
            cur.next = res;
            res = cur;
        }
        return res;
    }
}