leetcode题解-2. Add Two Numbers

题意：给你两个表示两个非负数字的链表。数字以相反的顺序存储，其节点包含单个数字。将这两个数字相加并将其作为一个链表返回。

例子：输入: (2 -> 4 -> 3) + (5 -> 6 -> 4)
输出: 7 -> 0 -> 8
解释：342 + 465 = 807.

分析：这道题比较简单，两个链表相加的问题，思路很明确，就是按照位数读下去，维护当前位和进位，时间复杂度是O(n),空间复杂度是O(1).

另外本题是链表部分的第一题，所以也附上链表部分的主函数和打印函数，以后的链表部分题目为了简介不在包含主函数。值得注意的是，因为本题定义了链表节点的数据结构，因此包含Solution.java和ListNode.java两部分的代码。以后为了精简博客，ListNode.java部分的代码也不再附上。

Solution.java

class Solution {
   public  ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1 == null && l2 == null) return null;
        ListNode point = new ListNode(0);
        ListNode head = point;
        int carry = 0;
        while(l1 != null && l2 != null){
        	int sum = carry + l1.val + l2.val;
        	point.next = new ListNode(sum % 10);
        	point = point.next;
        	carry = sum /10;
        	l1 = l1.next;
        	l2 = l2.next;
        }
        while(l1 != null){
        	int sum = carry + l1.val;
        	point.next = new ListNode(sum % 10);
        	point = point.next;
        	carry = sum /10;
        	l1 = l1.next;
        }
        while(l2 != null){
        	int sum = carry + l2.val;
        	point.next = new ListNode(sum % 10);
        	point = point.next;
        	carry = sum /10;
        	l2 = l2.next;
        }
        if(carry != 0){
        	point.next = new ListNode(carry);
        }
        
        
        return head.next;
    }
    public static void print(ListNode head){
    	while(head != null){
    		System.out.println(head.val);
    		head = head.next;
    	}
    	
    }
    public static void main(String[] args) {
		ListNode l1 = new ListNode(2);
		ListNode l2 = new ListNode(4);
		ListNode l3 = new ListNode(3);
		ListNode l4 = new ListNode(5);
		ListNode l5 = new ListNode(6);
		ListNode l6 = new ListNode(4);
		l1.next = l2;
		l2.next = l3;
		l4.next = l5;
		l5.next = l6;
		print(addTwoNumbers(l1, l4));
		
	}
}

ListNode.java

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

c++:

#include 

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        int carry = 0;
        ListNode *pointNode = new ListNode(0);
        ListNode *root = pointNode;
        while (l1 != NULL && l2 != NULL) {
            int sum = l1->val + l2->val + carry;
            int rem = sum % 10;
            carry = sum / 10;
            pointNode->next = new ListNode(rem);
            pointNode = pointNode->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        while (l1 != NULL) {
            int sum = l1->val + carry;
            int rem = sum % 10;
            carry = sum / 10;
            pointNode->next = new ListNode(rem);
            pointNode = pointNode->next;
            l1 = l1->next;
        }
        while (l2 != NULL) {
            int sum = l2->val + carry;
            int rem = sum % 10;
            carry = sum / 10;
            pointNode->next = new ListNode(rem);
            pointNode = pointNode->next;
            l2 = l2->next;
        }
        if (carry != 0) {
            pointNode->next = new ListNode(carry);
        }
        return root->next;
    }
};


int main() {

    ListNode* l1 = new ListNode(9);
    ListNode* l2 = new ListNode(1);
    ListNode* l3 = new ListNode(6);
    ListNode* l4 = new ListNode(0);
    l1->next = l2;
    l2->next = l3;
    Solution sl;
    ListNode* root = sl.addTwoNumbers(l1, l4);
    std::cout << "Hello, World!" << std::endl;
    return 0;
}