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array/1binary search/35search-insert-position/python/35. 搜索插入位置.ipynb

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# [35. 搜索插入位置](https://leetcode.cn/problems/search-insert-position)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## 题目描述\n",
+    "\n",
+    " 给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。\n",
+    "\n",
+    "请必须使用时间复杂度为 `O(log n)` 的算法。\n",
+    "\n",
+    "**示例 1:**\n",
+    "\n",
+    "```txt\n",
+    "输入: nums = [1,3,5,6], target = 5\n",
+    "输出: 2\n",
+    "```\n",
+    "\n",
+    "**示例 2:**\n",
+    "\n",
+    "```txt\n",
+    "输入: nums = [1,3,5,6], target = 2\n",
+    "输出: 1\n",
+    "```\n",
+    "\n",
+    "**示例 3:**\n",
+    "\n",
+    "```txt\n",
+    "输入: nums = [1,3,5,6], target = 7\n",
+    "输出: 4\n",
+    "```\n",
+    "\n",
+    "**提示:**\n",
+    "\n",
+    "-   `1 <= nums.length <= 10^4`\n",
+    "-   `-10^4 <= nums[i] <= 10^4`\n",
+    "-   `nums` 为 **无重复元素** 的 **升序** 排列数组\n",
+    "-   `-10^4 <= target <= 10^4`"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "---\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## 解答\n",
+    "\n",
+    " "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class Solution(object):\n",
+    "    def searchInsert(self, nums, target):\n",
+    "        \"\"\"\n",
+    "        :type nums: List[int]\n",
+    "        :type target: int\n",
+    "        :rtype: int\n",
+    "        \"\"\"\n",
+    "        left, right = 0, len(nums) - 1\n",
+    "        while left < right:\n",
+    "            mid = (left + right) // 2\n",
+    "            if target > nums[mid]:\n",
+    "                left = mid + 1\n",
+    "            elif target < nums[mid]:\n",
+    "                right = mid\n",
+    "            else:\n",
+    "                return mid \n",
+    "        return left if nums[left] >= target else left + 1  # 注意这里的返回值 注意溢出"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "id": "a1e23e83",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4\n"
+     ]
+    }
+   ],
+   "source": [
+    "test_nums = [1,3,5,6]\n",
+    "test_target = 7\n",
+    "solution_1 = Solution()\n",
+    "result_1 = solution_1.searchInsert(test_nums, test_target)\n",
+    "print(result_1) # 2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "92764a56",
+   "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": "3.8.10"
+  }
+ },
+ "nbformat": 5,
+ "nbformat_minor": 10
+}