From 57102d2044fb02320542130dc5a1d9c62a5d9019 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Thu, 22 Nov 2018 17:52:05 +0800
Subject: [PATCH] Create 2. Add Two Numbers.md

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 create mode 100644 2. Add Two Numbers.md

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+# [2. Add Two Numbers](https://leetcode.com/problems/add-two-numbers/)
+# 思路
+从前到尾不断将两个链表的元素相加就行了，这个过程中一直会用到一个代表进位的变量cin，另外注意处理两个链表不等长的情况。
+
+# C++
+``` C++
+class Solution {
+private:
+    int digit_add(const int a, const int b, int& cin){ // 先定义一个一位数字相加的函数，方便后面使用，注意是用引用的方式传入cin的
+        int sum = a + b + cin;
+        cin = sum / 10;
+        return sum % 10;
+    }
+public:
+    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
+        int cin = 0; // 进位
+        ListNode *root = new ListNode(digit_add(l1 -> val, l2 -> val, cin)); // root是最后要返回的指针
+        ListNode *last = root; // last指向当前结果链表中的最后一个元素
+        l1 = l1 -> next;
+        l2 = l2 -> next;
+        while(l1 && l2){  // 由前先后对两个链表的元素相加
+            last -> next = new ListNode(digit_add(l1 -> val, l2 -> val, cin));
+            last = last -> next;
+            l1 = l1 -> next;
+            l2 = l2 -> next;
+        }
+        if(l2) l1 = l2; // 若l2长一些，将l1指向l2，后续就只对l1处理就行了.
+        while(l1 && cin > 0){
+            last -> next = new ListNode(digit_add(l1 -> val, 0, cin));
+            last = last -> next;
+            l1 = l1 -> next;
+        }
+        if(l1) last -> next = l1; // cin = 0，直接将l1后续的所有节点接到last后面就行了
+        else{   // l1 == NULL， cin >= 0
+            if(cin > 0){
+                last -> next = new ListNode(cin);
+                last = last -> next;    
+            }
+            last -> next = NULL;
+        }
+        return root;
+    }
+};
+```