LeetCode刷题系列——Add Two Numbers

题目链接

 

　　这个题目很简单，归并而已，好久没练编程，居然忘了在使用自定义类型前，要进行初始化（new操作）。

class ListNode{
    int val;
    ListNode next;
    ListNode(int x){
        val = x;
    }
}
//在使用之前ListNode得next变量前需要进行初始化：
ListNode node = new ListNode(1);
a = node.next;
a = new ListNode(2);
//上面是不能通过node找到刚刚初始化的a, 只有通过下面这种方式才能找到next，形成一条链
node.next = new ListNode(3);
a = node.next; //此时对a操作才能，才能形成对一条链的操作

　　以前我认为归并需要分成3部分，while()直到node1和node2其中一个为空，然后再while()非空的那个node，但是不同场景应用方式也不同，如果遇到归并相加问题，就不用那么麻烦，代码对比一下就知道了。

public class Solution {
    
    public static void main(String[] args) {
        int[] data1 = new int[] {1};
        int[] data2 = new int[] {9,9};
        ListNode node1 = new ListNode(data1[0]);
        ListNode node2 = new ListNode(data2[0]);
        ListNode tmp = node1;
        for(int i = 1; i < data1.length; i ++) {
            tmp.next = new ListNode(data1[i]);
            tmp = tmp.next;
        }
        tmp = node2;
        for(int i = 1; i < data2.length; i ++) {
            tmp.next = new ListNode(data2[i]);
            tmp = tmp.next;
        }
        Solution s = new Solution();
        tmp = s.addTwoNumbers(node1, node2);
        while(tmp != null) {
            System.out.format("%d ", tmp.val);
            tmp = tmp.next;
        }
    }
    
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode cur_node = head;
        int carry = 0;
        while(l1 != null || l2 != null) {
            int x = (l1 == null) ? 0 : l1.val;
            int y = (l2 == null) ? 0 : l2.val;
            int val = x + y + carry;
            carry = val / 10;
            cur_node.next = new ListNode(val % 10);
            cur_node = cur_node.next;
            if(l1 != null)
                l1 = l1.next;
            if(l2 != null)
                l2 = l2.next;
        }
        if(carry > 0) {
            cur_node.next = new ListNode(1);
        }
        return head.next;
    }
    
    private class HelpMergeReturn{
        int flag;
        ListNode node;
        HelpMergeReturn(int f, ListNode n){
            flag = f;
            node = n;
        }
    }
    
    private HelpMergeReturn helpMerge(ListNode node, ListNode l, int flag) {
        while(l != null) {
            int val = l.val + flag;
            flag = val / 10;
            node.next = new ListNode(val % 10);
            node = node.next;
            l = l.next;
        }
        return new HelpMergeReturn(flag, node);
    }
    
    public ListNode addTwoNumbers1(ListNode l1, ListNode l2) {
        int flag = 0;
        ListNode result = null;
        ListNode tmp_node = null;
        while(true) {
            if(l1 == null || l2 == null)
                break;
            int val = l1.val + l2.val + flag;
            flag = val / 10;
            if(tmp_node == null) {
                tmp_node = new ListNode(val % 10);
                result = tmp_node;
            }
            else {
                tmp_node.next = new ListNode(val % 10);
                tmp_node = tmp_node.next;
            }
            l1 = l1.next;
            l2 = l2.next;
        }
        HelpMergeReturn hm = null;
        if(l1 != null) {
            hm = helpMerge(tmp_node, l1, flag);
        }
        if(l2 != null) {
            hm = helpMerge(tmp_node, l2, flag);
        }
        if(hm != null) {
            tmp_node = hm.node;
            flag = hm.flag;
        }
        if(flag != 0) {
            tmp_node.next = new ListNode(flag);
        }
        return result;
    }
}