Add Two Numbers

Jack-Cherish · Jack-Cherish · commit 7e48f7240f63 · 2017-10-12T09:18:00.000+08:00
diff --git a/Linked List/Medium/2.Add Two Numbers.md b/Linked List/Medium/2.Add Two Numbers.md
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+# 题目
+
+给定两个非空链表，结点元素为非负整数，这些数字从小到大排列。将对应结点的两个数字相加，并返回一个新链表。
+
+假设两个链表不以0打头。
+
+**举例：**
+
+``` stylus
+Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
+Output: 7 -> 0 -> 8
+```
+
+# 思路
+
+对应位置元素相加，注意进位即可。使用divmod函数进行计算即可，很简单。
+
+# 代码
+
+**Python：**
+
+``` python
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution(object):
+    def addTwoNumbers(self, l1, l2):
+        """
+        :type l1: ListNode
+        :type l2: ListNode
+        :rtype: ListNode
+        """
+        extra = 0
+        root = n = ListNode(0)
+        while l1 or l2 or extra:
+            v1 = v2 = 0
+            if l1:
+                v1 = l1.val
+                l1 = l1.next
+            if l2:
+                v2 = l2.val
+                l2 = l2.next
+            extra, val = divmod(v1 + v2 + extra, 10)
+            n.next = ListNode(val)
+            n = n.next
+        return root.next
+```
+
+**C++：**
+
+``` cpp
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
+        ListNode preHead(0), *p = &preHead;
+        int extra = 0;
+        while (l1 || l2 || extra) {
+            int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;
+            extra = sum / 10;
+            p->next = new ListNode(sum % 10);
+            p = p->next;
+            l1 = l1 ? l1->next : l1;
+            l2 = l2 ? l2->next : l2;
+        }
+        return preHead.next;
+    }
+};
+```
+
+
