[LeetCode] 445. 两数相加 II

Posted 2021-03-01 怕什么

自己想的太复杂了。。。。。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode pre=null;
        ListNode head=null;
        ListNode l1node=l1,l2node=l2;
        pre.next=head;
        int l1len=0;
        int l2len=0;
        whille(l1node!=null){
            l1len++;
            l1node=l1node.next;
        }
        whille(l2node!=null){
            l2len++;
            l2node=l2node.next;
        }
        if(l1len>l2len){
            int m=l1len-l2len;
            while(m!=0){
                head=l1;
                pre=head;
                head=head.next;
                l1=l1.next; 
                m--; 
            }
        }else{
            int m=l2len-l1len;
            while(m!=0){
                head=l1;
                pre=head;
                head=head.next;
                l2=l2.next; 
                m--; 
            }
        }
        while(l1!=null){
            int c1=l1==null?0:c1.val;
            int c2=l2==null?0:c2.val;
            int sum=(c1+c2)%10;
            int carry=(c1+c2)/10;
            if(carry==1) pre+=1;
            head=sum;
            pre=sum;
            head=head.next;
        }
    }
}

View Code

然而可以用栈放进去，出来就是数字从低位到高位了。

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 
        Stack<Integer> stack1 = new Stack<>();
        Stack<Integer> stack2 = new Stack<>();
        while (l1 != null) {
            stack1.push(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            stack2.push(l2.val);
            l2 = l2.next;
        }
        
        int carry = 0;
        ListNode head = null;
        while (!stack1.isEmpty() || !stack2.isEmpty() || carry > 0) {
            int sum = carry;
            sum += stack1.isEmpty()? 0: stack1.pop();
            sum += stack2.isEmpty()? 0: stack2.pop();
            ListNode node = new ListNode(sum % 10);
            node.next = head;
            head = node;
            carry = sum / 10;
        }
        return head;
    }
}

或者是直接翻转链表，然后按照第2题那样的方法去做更快

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummyHead = new ListNode(0);
        ListNode p = reverseList(l1);
        ListNode q = reverseList(l2);
        ListNode cur = 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 = x + y + carry;
            carry = sum/10;
            cur.next = new ListNode(sum%10);
            cur = cur.next;
            if (p != null){
                p = p.next;
            }
            if (q != null){
                q = q.next;
            }
        }
        if (carry == 1){
            cur.next = new ListNode(carry);
        }
        return reverseList(dummyHead.next);
    }
    public static ListNode reverseList(ListNode cur) {
        ListNode pre = null;
        while (cur != null) {
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
}