Add Two Numbers

include/s2_add_two_numbers.hpp

@@ -0,0 +1,31 @@
+#ifndef S2_ADD_TWO_NUMBERS_HPP
+#define S2_ADD_TWO_NUMBERS_HPP
+
+struct ListNode {
+  int val;
+  ListNode* next;
+  ListNode() : val(0), next(nullptr) {}
+  ListNode(int x) : val(x), next(nullptr) {}
+  ListNode(int x, ListNode* next) : val(x), next(next) {}
+};
+
+class Solution {
+ public:
+  /**
+   * @brief Add Two Numbers
+   *
+   * You are given two **non-empty** linked lists representing two non-negative
+   * integers. The digits are stored in **reverse order**, and each of their
+   * nodes contains a single digit. Add the two numbers and return the sum as a
+   * linked list.
+   *
+   * You may assume the two numbers do not contain any leading zero, except the
+   * number 0 itself.
+   *
+   * @param l1 the first linked list
+   * @param l2 the second linked list
+   */
+  ListNode* addTwoNumbers(ListNode* l1, ListNode* l2);
+};
+
+#endif

src/s2_add_two_numbers.cpp

@@ -0,0 +1,23 @@
+#include "s2_add_two_numbers.hpp"
+
+ListNode* Solution::addTwoNumbers(ListNode* l1, ListNode* l2) {
+  ListNode *l_begin = new ListNode((l1->val + l2->val) % 10);
+  ListNode *l_end = l_begin;
+  int carry{ (l1->val + l2->val) >= 10 ? 1 : 0 };
+  while (l1->next != nullptr || l2->next != nullptr || carry != 0) {
+    if (l1->next == nullptr && l2->next != nullptr) {
+      l1->next = new ListNode(0);
+    } else if (l1->next != nullptr && l2->next == nullptr) {
+      l2->next = new ListNode(0);
+    } else if (l1->next == nullptr && l2->next == nullptr && carry != 0) {
+      l1->next = new ListNode(0);
+      l2->next = new ListNode(0);
+    }
+    l1 = l1->next;
+    l2 = l2->next;
+    l_end->next = new ListNode((l1->val + l2->val + carry) % 10);
+    l_end = l_end->next;
+    carry = (l1->val + l2->val + carry) >= 10 ? 1 : 0;
+  }
+  return l_begin;
+}

tests/s2_add_two_numbers.cpp

@@ -0,0 +1,53 @@
+#include "s2_add_two_numbers.hpp"
+
+#include <gtest/gtest.h>
+
+TEST(Problem2, Case1) {
+  ListNode *l1 =
+    new ListNode(2,
+      new ListNode(4,
+        new ListNode(3)));
+  ListNode *l2 =
+    new ListNode(5,
+        new ListNode(6,
+          new ListNode(4)));
+  Solution solution;
+  ListNode *l = solution.addTwoNumbers(l1, l2);
+  EXPECT_EQ(l->val, 7);
+  EXPECT_EQ(l->next->val, 0);
+  EXPECT_EQ(l->next->next->val, 8);
+}
+
+TEST(Problem2, Case2) {
+  ListNode *l1 = new ListNode(0);
+  ListNode *l2 = new ListNode(0);
+  Solution solution;
+  ListNode *l = solution.addTwoNumbers(l1, l2);
+  EXPECT_EQ(l->val, 0);
+}
+
+TEST(Problem2, Case3) {
+  ListNode *l1 =
+    new ListNode(9,
+        new ListNode(9,
+          new ListNode(9,
+            new ListNode(9,
+              new ListNode(9,
+                new ListNode(9,
+                  new ListNode(9)))))));
+  ListNode *l2 =
+    new ListNode(9,
+        new ListNode(9,
+          new ListNode(9,
+            new ListNode(9))));
+  Solution solution;
+  ListNode *l = solution.addTwoNumbers(l1, l2);
+  EXPECT_EQ(l->val, 8);
+  EXPECT_EQ(l->next->val, 9);
+  EXPECT_EQ(l->next->next->val, 9);
+  EXPECT_EQ(l->next->next->next->val, 9);
+  EXPECT_EQ(l->next->next->next->next->val, 0);
+  EXPECT_EQ(l->next->next->next->next->next->val, 0);
+  EXPECT_EQ(l->next->next->next->next->next->next->val, 0);
+  EXPECT_EQ(l->next->next->next->next->next->next->next->val, 1);
+}