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Author: assassint2017

Add Two Numbers II/Add_Two_Numbers_II.cpp

@@ -0,0 +1,157 @@
+// 其实题目中限制了不能对原始链表进行修改反倒是好事，这样就容易了很多
+// 实践之后发现把问题想的太简单了，这样写最大的问题就是很容易溢出
+// 1560 / 1563 test cases passed.
+
+/**
+ * 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) 
+    {
+        long long sum1 = 0, sum2 = 0;
+        
+        ListNode* ptr1 = l1;
+        ListNode* ptr2 = l2;
+        
+        while (ptr1)
+        {
+            sum1 *= 10;
+            sum1 += ptr1->val;
+            ptr1 = ptr1->next;
+        }
+        
+        while (ptr2)
+        {
+            sum2 *= 10;
+            sum2 += ptr2->val;
+            ptr2 = ptr2->next;
+        }
+        
+        long long res = sum1 + sum2;
+
+        // 根据计算出来的和，构建链表
+        ListNode dummyHead(0);
+
+        while (res != 0)
+        {
+            int val = res % 10;
+
+            ListNode* node = new ListNode(val);
+            ListNode* temp = dummyHead.next;
+
+            dummyHead.next = node;
+            node->next = temp;
+
+            res /= 10;
+        }
+        return dummyHead.next;
+    }
+};
+
+
+// 所以使用堆栈来模拟加法过程
+// Runtime: 24 ms, faster than 74.03% of C++ online submissions for Add Two Numbers II.
+// Memory Usage: 13 MB, less than 51.85% of C++ online submissions for Add Two Numbers II.
+
+/**
+ * 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) 
+    {
+        stack<ListNode*> stack1, stack2;
+
+        ListNode* ptr1 = l1;
+        ListNode* ptr2 = l2;
+
+        while (ptr1)
+        {
+            stack1.push(ptr1);
+            ptr1 = ptr1->next;
+        }
+
+        while (ptr2)
+        {
+            stack2.push(ptr2);
+            ptr2 = ptr2->next;
+        }
+
+        int base = 0, carry = 0;
+        ListNode dummyHead(0);
+        while (!stack1.empty() && !stack2.empty())
+        {
+            ptr1 = stack1.top();
+            stack1.pop();
+
+            ptr2 = stack2.top();
+            stack2.pop();
+
+            base = (ptr1->val + ptr2->val + carry) % 10;
+            carry = (ptr1->val + ptr2->val + carry) / 10;
+            
+            ListNode* node = new ListNode(base);
+            ListNode* temp = dummyHead.next;
+
+            dummyHead.next = node;
+            node->next = temp;
+        }
+
+        while (!stack1.empty())
+        {
+            ptr1 = stack1.top();
+            stack1.pop();
+
+            base = (ptr1->val + carry) % 10;
+            carry = (ptr1->val + carry) / 10;
+            
+            ListNode* node = new ListNode(base);
+            ListNode* temp = dummyHead.next;
+
+            dummyHead.next = node;
+            node->next = temp;
+        }
+
+        while (!stack2.empty())
+        {
+            ptr2 = stack2.top();
+            stack2.pop();
+
+            base = (ptr2->val + carry) % 10;
+            carry = (ptr2->val + carry) / 10;
+            
+            ListNode* node = new ListNode(base);
+            ListNode* temp = dummyHead.next;
+
+            dummyHead.next = node;
+            node->next = temp;
+        }
+
+        // 这里要特别注意一下 [5]+[5]=[1][0]
+        if (carry != 0)
+        {
+            ListNode* node = new ListNode(carry);
+            ListNode* temp = dummyHead.next;
+
+            dummyHead.next = node;
+            node->next = temp;
+        }
+
+        return dummyHead.next;
+    }
+};