homework

2023/homework/XuWeiZhang-UCAS/LeetcodeHW.ipynb

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "434. Number of Segments in a String"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class Solution(object):\n",
+    "    def countSegments(self, s):\n",
+    "        \"\"\"\n",
+    "        :type s: str\n",
+    "        :rtype: int\n",
+    "        \"\"\"\n",
+    "        return len(s.split()) #运用内建的split()函数"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "852. Peak Index in a Mountain Array"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class Solution(object):\n",
+    "    def peakIndexInMountainArray(self, arr):\n",
+    "        \"\"\"\n",
+    "        :type arr: List[int]\n",
+    "        :rtype: int\n",
+    "        \"\"\"\n",
+    "        # 遍历，对于每一个元素，判断它是否大于它前面和后面的元素，如果是，则返回当前索引\n",
+    "        for i in range(1, len(arr)-1):\n",
+    "            if arr[i] > arr[i-1] and arr[i] > arr[i+1]:\n",
+    "                return i"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "1784. Check if Binary String Has at Most One Segment of Ones\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class Solution(object):\n",
+    "    def checkZeroOnes(self, s):\n",
+    "        \"\"\"\n",
+    "        :type s: str\n",
+    "        :rtype: bool\n",
+    "        \"\"\"\n",
+    "        x = s.split('0')\n",
+    "        y = s.split('1') # 用split函数把字符串的各个独立部分分割\n",
+    "\n",
+    "        mx0,mx1 = 0,0 #初始化0和1的长度参数\n",
+    "\n",
+    "        # 遍历两组字符串的组，用max函数获得最长的字符串\n",
+    "        for i in x:\n",
+    "            mx0 = max(mx0, len(i))\n",
+    "        for i in y:\n",
+    "            mx1 = max(mx1, len(i))\n",
+    "        return (mx0 > mx1)# 比较长度得到结果"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "1869. Longer Contiguous Segments of Ones than Zeros"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class Solution(object):\n",
+    "    def checkOnesSegment(self, s):\n",
+    "        \"\"\"\n",
+    "        :type s: str\n",
+    "        :rtype: bool\n",
+    "        \"\"\"\n",
+    "        count = 0 # 计数初始化\n",
+    "        x = s.split(\"0\") # 以0分割字符串\n",
+    "        # 遍历分割后的数组，计数\n",
+    "        for i in x:\n",
+    "            if i:\n",
+    "                count += 1\n",
+    "        return count == 1 # 判断个数"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "162. Find Peak Element"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class Solution(object):\n",
+    "    def findPeakElement(self, nums):\n",
+    "        \"\"\"\n",
+    "        :type nums: List[int]\n",
+    "        :rtype: int\n",
+    "        \"\"\"\n",
+    "        max = nums[0]\n",
+    "        peak = 0 # 初始化参数\n",
+    "        for i in range(len(nums)): # 循环判断最大值并获得此时序号\n",
+    "            if nums[i] > max:\n",
+    "                max = nums[i]\n",
+    "                peak = i\n",
+    "\n",
+    "        return peak # 返回序号"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "ictp-ap",
+   "language": "python",
+   "name": "ictp-ap"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.13"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}

2023/homework/XuWeiZhang-UCAS/numpy_submit.txt

@@ -0,0 +1,10 @@
+C
+D
+B
+E
+C
+C
+E
+C
+C
+E

2023/homework/XuWeiZhang-UCAS/pandas_submit.txt

@@ -0,0 +1,12 @@
+B
+D
+C
+A
+B
+D
+A
+C
+C
+B
+B
+B

2023/homework/XuWeiZhang-UCAS/python_submit.txt

@@ -0,0 +1,108 @@
+C
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