Add leetcode exercises

algorithms/001-two-sum.ipynb

@@ -0,0 +1,109 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "common-configuration",
+   "metadata": {},
+   "source": [
+    "给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。\n",
+    "你可以假设每种输入只会对应一个答案。但是，数组中同一个元素不能使用两遍。\n",
+    "\n",
+    "#### 示例:\n",
+    "- 给定 `nums = [2, 7, 11, 15]`, `target = 9`\n",
+    "- 因为 `nums[0] + nums[1] = 2 + 7 = 9`\n",
+    "- 所以返回 `[0, 1]`"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "black-discovery",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def solve(nums, target):\n",
+    "    candidates = {}\n",
+    "    for i, n in enumerate(nums):\n",
+    "        candidates[target - n] = i\n",
+    "    for i, n in enumerate(nums):\n",
+    "        if n in candidates and i != candidates[n]:\n",
+    "            return [i, candidates[n]]\n",
+    "    return None"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "id": "refined-palestinian",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "[0, 1]"
+      ]
+     },
+     "execution_count": 5,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "solve([2, 7, 11, 15], 9)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "id": "early-reader",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1, 3]\n",
+      "[1, 2]\n",
+      "[0, 3]\n",
+      "[2, 4]\n"
+     ]
+    }
+   ],
+   "source": [
+    "print(solve([11, 7, 15, 2], 9))\n",
+    "print(solve([3, 2, 4], 6))\n",
+    "print(solve([0, 4, 3, 0], 0))\n",
+    "print(solve([-1, -2, -3, -4, -5], -8))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "little-demand",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.8.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}

algorithms/001-two-sum.py

@@ -0,0 +1,29 @@
+"""
+给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。
+
+你可以假设每种输入只会对应一个答案。但是，数组中同一个元素不能使用两遍。
+
+#### 示例:
+
+给定 nums = [2, 7, 11, 15], target = 9
+因为 nums[0] + nums[1] = 2 + 7 = 9
+所以返回 [0, 1]
+"""
+
+
+def two_sum(nums, target: int):
+    candidates = {}
+    for i, n in enumerate(nums):
+        # if target >= n:
+        candidates[target - n] = i
+    for i, n in enumerate(nums):
+        if n in candidates and i != candidates[n]:
+            return [i, candidates[n]]
+
+
+if __name__ == '__main__':
+    print(two_sum([2, 7, 11, 15], 9))
+    print(two_sum([11, 7, 15, 2], 9))
+    print(two_sum([3, 2, 4], 6))
+    print(two_sum([0, 4, 3, 0], 0))
+    print(two_sum([-1, -2, -3, -4, -5], -8))

algorithms/002-add-two-numbers.ipynb

@@ -0,0 +1,107 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "stainless-cross",
+   "metadata": {},
+   "source": [
+    "### [两个链表生成相加链表](https://leetcode-cn.com/problems/add-two-numbers/)\n",
+    "\n",
+    "给你两个**非空**的链表，表示两个非负的整数。它们每位数字都是按照**逆序**的方式存储的，并且每个节点只能存储**一位**数字。\n",
+    "\n",
+    "请你将两个数相加，并以相同形式返回一个表示和的链表。\n",
+    "\n",
+    "你可以假设除了数字`0`之外，这两个数都不会以`0`开头。\n",
+    "\n",
+    "#### 示例 1：\n",
+    "```\n",
+    "输入：l1 = [2,4,3], l2 = [5,6,4]\n",
+    "输出：[7,0,8]\n",
+    "解释：342 + 465 = 807.\n",
+    "```\n",
+    "\n",
+    "#### 示例 2：\n",
+    "```\n",
+    "输入：l1 = [0], l2 = [0]\n",
+    "输出：[0]\n",
+    "```\n",
+    "\n",
+    "#### 示例 3：\n",
+    "```\n",
+    "输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]\n",
+    "输出：[8,9,9,9,0,0,0,1]\n",
+    "```\n",
+    "#### 提示：\n",
+    "- 每个链表中的节点数在范围`[1, 100]`内\n",
+    "- `0 <= Node.val <= 9`\n",
+    "- 题目数据保证列表表示的数字不含前导零"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "tested-consideration",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class ListNode:\n",
+    "    def __init__(self, val=0, next=None):\n",
+    "        self.val = val\n",
+    "        self.next = next\n",
+    "\n",
+    "class Solution:\n",
+    "    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:\n",
+    "        h1, h2, ans = l1, l2, ListNode()\n",
+    "        carry, curr = 0, ans\n",
+    "        while h1 or h2:\n",
+    "            h1val = h1.val if h1 else 0\n",
+    "            h2val = h2.val if h2 else 0\n",
+    "            value = h1val + h2val + carry\n",
+    "            if value >= 10:\n",
+    "                carry = 1\n",
+    "                value = value % 10\n",
+    "            else:\n",
+    "                carry = 0\n",
+    "            curr.next = ListNode(value)\n",
+    "            if h1 is not None:\n",
+    "                h1 = h1.next\n",
+    "            if h2 is not None:\n",
+    "                h2 = h2.next\n",
+    "            curr = curr.next\n",
+    "        \n",
+    "        if carry > 0:\n",
+    "            curr.next = ListNode(carry)\n",
+    "        return ans.next"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "joint-retro",
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.8.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}

algorithms/002-add-two-numbers.py

@@ -0,0 +1,112 @@
+"""
+两数相加
+
+给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。
+
+如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。
+
+您可以假设除了数字 0 之外，这两个数都不会以 0 开头。
+
+示例：
+
+输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
+输出：7 -> 0 -> 8
+原因：342 + 465 = 807
+"""
+
+
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+
+# def add_two_numbers(l1: ListNode, l2: ListNode) -> ListNode:
+#     p, q = l1, l2
+#     curr = head = ListNode(None)
+#     carry = 0
+#     while p is not None and q is not None:
+#         val = p.val + q.val + carry
+#         if val >= 10:
+#             val = val % 10
+#             carry = 1
+#         else:
+#             carry = 0
+#
+#         curr.next = ListNode(val)
+#         curr = curr.next
+#         p, q = p.next, q.next
+#
+#     if carry == 0:
+#         if p is not None:
+#             curr.next = p
+#         elif q is not None:
+#             curr.next = q
+#     else:
+#         while p is not None:
+#             val = p.val + carry
+#             if val >= 10:
+#                 val = val % 10
+#                 curr.next = ListNode(val)
+#                 curr = curr.next
+#             else:
+#                 curr.next = ListNode(val)
+#                 curr = curr.next
+#                 curr.next = p.next
+#                 carry = 0
+#                 break
+#             p = p.next
+#
+#         while q is not None:
+#             val = q.val + carry
+#             if val >= 10:
+#                 val = val % 10
+#                 curr.next = ListNode(val)
+#                 curr = curr.next
+#             else:
+#                 curr.next = ListNode(val)
+#                 curr = curr.next
+#                 curr.next = q.next
+#                 carry = 0
+#                 break
+#             q = q.next
+#
+#         if carry == 1:
+#             curr.next = ListNode(1)
+#
+#     return head.next
+
+
+def add_two_numbers(l1: ListNode, l2: ListNode) -> ListNode:
+    p, q = l1, l2
+    curr = head = ListNode(None)
+    carry = 0
+    while p is not None or q is not None:
+        pval = p.val if p is not None else 0
+        qval = q.val if q is not None else 0
+        val = pval + qval + carry
+        if val >= 10:
+            val = val % 10
+            carry = 1
+        else:
+            carry = 0
+
+        curr.next = ListNode(val)
+        curr = curr.next
+        if p is not None:
+            p = p.next
+        if q is not None:
+            q = q.next
+
+    if carry == 1:
+        curr.next = ListNode(1)
+
+    return head.next
+
+
+if __name__ == '__main__':
+    l1 = ListNode(9)
+    l1.next = ListNode(8)
+    l2 = ListNode(1)
+    print(add_two_numbers(l1, l2))
+