leetcode-2-两数相加

Posted 2022-03-10 nxzblogs

 

问题：

 

 

package com.nxz.blog.otherTest;

public class TestLeetCode 

    public static void main(String[] args) 

        TestLeetCode t = new TestLeetCode();
        ListNode l1 = new ListNode(2);
        ListNode l2 = new ListNode(4);
        ListNode l3 = new ListNode(3);
        ListNode r1 = new ListNode(5);
        ListNode r2 = new ListNode(6);
        ListNode r3 = new ListNode(4);
        l1.next = l2;
        l2.next = l3;
        r1.next = r2;
        r2.next = r3;
        ListNode listNode = t.addTwoNumbers(l1, r1);

        do 
            System.out.println(listNode.val);
         while ((listNode = listNode.next )!= null);
    


    /**
     * 循环 l1 和 l2 两个listnode，当l1 或 l2 不为null的时候，将两个数相加，大于10时进行处理（保存进一位），否则视为0和另一个值相加
     *
     * @param l1
     * @param l2
     * @return
     */
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) 
　　　　　//哑结点，最终返回的节点就是该节点的下一个节点
        ListNode dummyHead = new ListNode(-1);
        //暂存变量
        ListNode p = l1, q = l2, curr = dummyHead;
        //大于10时的暂存值
        int carry = 0;

        //当p或q不为null是，将值相加，此时需要处理，其中一个为null的情况
        while (p != null || q != null) 
            //只要p或q为nul时，将val值默认为0，和另一个值相加
            int x = p != null ? p.val : 0;
            int y = q != null ? q.val : 0;

            int sum = carry + x + y;
            //重新设置进一变量
            carry = sum / 10;
            curr.next = new ListNode(sum % 10);
            //重新设置当前节点，p，q节点，以便再一次循环
            curr = curr.next;
            if (p != null) 
                p = p.next;
            
            if (q != null) 
                q = q.next;
            
        

        //最后处理最高为的carry的值
        if (carry > 0) 
            curr.next = new ListNode(carry);
        

        return dummyHead.next;

    

    public static class ListNode 
        int val;
        ListNode next;

        ListNode(int x) 
            val = x;
        
    

 

 

 进阶----->>>>

两数相加：节点为正序的情况

package com.nxz.blog.otherTest;

import java.util.Stack;

public class TestLeetCode 

    public static void main(String[] args) 

        TestLeetCode t = new TestLeetCode();
        ListNode l1 = new ListNode(2);
        ListNode l2 = new ListNode(4);
        ListNode l3 = new ListNode(3);
        ListNode r1 = new ListNode(5);
        ListNode r2 = new ListNode(6);
        ListNode r3 = new ListNode(4);
        l1.next = l2;
        l2.next = l3;
        r1.next = r2;
        r2.next = r3;
        ListNode listNode = t.addTwoNumbers(l1, r1);

        do 
            System.out.println(listNode.val);
         while ((listNode = listNode.next) != null);
    


    /**
     * 利用栈这种数据结构（先进后出），这样就可以将正序的节点转换为倒叙的节点了，这样依次从两个栈中取出值（顺序就是个、十、百。。。）
     * 这样就转换为之前的那种方式了
     *
     * @param l1
     * @param l2
     * @return
     */
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) 
        Stack<ListNode> stack1 = new Stack<>();
        Stack<ListNode> stack2 = new Stack<>();
        ListNode dummyHead = new ListNode(-1);
        ListNode p = l1, q = l2, head = dummyHead;
        while (p != null) 
            stack1.add(p);
            p = p.next;
        
        while (q != null) 
            stack2.add(q);
            q = q.next;
        

        p = stack1.pop();
        q = stack2.pop();


        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;
            ListNode last = head.next;
            head.next = new ListNode(sum % 10);
            head.next.next = last;

            p = stack1.isEmpty() ? null : stack1.pop();
            q = stack2.isEmpty() ? null : stack2.pop();

        

        if (carry > 0) 
            ListNode last = head.next;
            head.next = new ListNode(carry);
            head.next.next = last;
        

        return dummyHead.next;
    

    public static class ListNode 
        int val;
        ListNode next;

        ListNode(int x) 
            val = x;