LeetCode cpp最新中文题解.pdf

LeetCode cpp最新中文题解.pdfLeetCode cpp最新中文题解.pdfLeetCode cpp最新中文题解.pdf目录3.4 Add binary615.1.5 Binary Tree Level Or-3.5 Longest Palindromic Substring. 62der traversal il3.6 Regular Expression Matching665.1.6 Binary Tree Zigzag3.7 Wildcard Matching67Level Order traversal. 963.8 Longest Common Prefix5.1.7 Recover Binary Search3. 9 Valid Number70Tree983.10 Integer to roman725. 1. 8 Same Tree3. 11 Roman to Integer735.1.9 Symmetric Tree1003.12 Count and Say745.1.10 Balanced Binary Tree.. 1023. 13 Anagrams755.1.11 Flatten Binary Tree to3. 14 Simplify Path76Linked List1033. 15 Length of Last Word775.1. 12 Populating Next RightPointers in each node ii 105第4章栈和队列7952二叉树的构建10641栈795.2.1 Construct Binary Tree4Valid Parentheses79from Preorder and In4.1.2 Longest valid Parenorder Traversa106theses805.2.2 Construct Binary Tree4.1.3 Largest Rectangle infrom Inorder and posHistogram82torder Traversal1074.1.4 Evaluate reverse pol-53二叉查找树108ish notation845.3. 1 Unique Binary Search4.2队列85Trees.1085.3.2 Unique Binary Search第5章树86Trees li.1105.1二叉树的遍历865.3.3 Validate Binary Search5.1.1 Binary Tree PreorderTreeTraversal865.3. 4 Convert Sorted array to5.1.2 Binary Tree InorderBinary Search Tree...112Traversal885.3.5 Convert Sorted List to5.1. 3 Binary Tree PostorderBinary Search Tree113Traversal9054二叉树的递归.1145. 1. 4 Binary Tree Level Or5.4.1 Minimum Depth of Bider traversal)2nary lree115目录5.4.2 Maximum Depth of Bi8.3.,2重新实现 next permunary Tree116tation1425.4.3 Path Sum11783.3递归.1435.4 4 Path Sum il118 8.4 Permutations II1445.4.5 Binary Tree Maximum8.4.1 next permutation... 144Path Suum11984.2重新实现 next permu5.4.6 Populating Next Righttation144Pointers in each node 120843递归1445.4.7 Sum Root to Leaf num8.5 Combinations146bers2185.1递归1468.5.2迭代147第6章排序1238.6 Letter Combinations of a phone6.1 Merge Sorted Array123umber1476.2 Merge Two Sorted Lists12486.1递归1486.3 Merge k Sorted Lists124862迭代96.4 Insertion Sort List125第9章广度优先搜索1506.5 Sort list1269.1 Word Ladder1506.6 First Missing Positive1279.2 Word Ladder il1546.7 Sort Colors1289.3 Surrounded regions162第7章查找94小结16413194.l适用场景1647.1 Search for a range131942思考的步骤7.2 Search Insert Position.13294.3代码模板1657. 3 Search a 2D Matrix133第10章深度优先搜索173第8章暴力枚举法13510.1 Palindrome Partitioning1738.1 Subsets13510.2 Unique Paths1768.1.1递归1350.2.1深搜1768.1.2迭代.1371022备忘录法.1768.2 Subsets il13810.23动规177821递归13810.24数学公式1788.2.2迭代.14110.3 Unique Paths Il1798. 3 Permutations14210.3.1备忘录法1798.3.1 next permutation14210.3.2动规.180目录10.4 N-Queens1813.4 Maximal rectangle21310.5 N-Queens II18413.5 Best Time to Buy and Sell Stock10.6 Restore ip addresses186.21410.7 Combination Sum18813.6 Interleaving String21510.8 Combination Sum Il18913.7 Scramble String21710.9 Generate Parentheses.19013. 8 Minimum Path Sum.22210.10 Sudoku solver19213.9 Edit Distance22410.11 Word Search.19313. 10 Decode Ways.22610.12小结19513. 11 Distinct sub22710.12.1适用场景19513. 12 Word Break22810.122思考的步骤19513 13 Word Break il2300.12.3代码模板197第14章图23210.12.4深搜与回溯法的区別.19714. 1 Clone Graph23210.12.5深搜与递归的区别..197第15章细节实现题235第11章分治法19915.1 Reverse Integer2351.1 Pow(x, n)19915.2 Palindrome Number.23611. 2 Sqrt(x)20015.3 Insert Interval237第12章贪心法20115.4 Merge Intervals23812.1 Jump game20115.5 Minimum Window Substring23912.2 Jump game II15.6 Multiply Strings24112. 3 Best Time to buy and Sell stock 20415.7 Substring with Concatenation12. 4 Best Time to buy and sell stock l205of all words24412. 5 Longest Substring Without re15.8 Pascal,s Triangle245peating Characters20615.9 Pascals Triangle Il24612.6 Container with Most Water.. 207 15.10 Spiral Matrix24715.11 Spiral matrix II248第13章动态规划20915.12 ZigZag Conversion25013. 1 Triangle20915.13 Divide Two Integers25113.2 Maximum Subarray15. 14 Text Justification25313.3 Palindrome Partitioning II1215.15 Max Points on a line255目录第1章编程技巧在判断两个浮点数a和b是否相等时,不要用a==b,应该判断二者之差的绝对值fabs(a-b)是否小于某个阈值,例如1e-9。判断一个整数是否是为奇数,用x%2!=0,不要用x%2=1,因为ⅹ可能是负用char的值作为数组下标(例如,统计字符串中每个字符岀现的次数),要考虑到char可能是负数。有的人考虑到了,先强制转型为 unsigned int再用作下标,这仍然是错的。正确的做法是,先强制转型为 unsigned char,再用作下标。这涉及C十整型提升的规则,就不详述了。以下是关于STL使用技巧的,很多条款来自《 Effective STL》这本书。vector和 string优先于动态分配的数组首先,在性能上,由于 vector能够保证连续内存,因此一旦分配了后,它的性能跟原始数组相当;其次,如果用new,意味着你要确保后面进行孓 delete,一旦忘记了,就会出现BUG,且这样需要都写一行 delete,代码不够短再次,声明多维数组的话,只能一个一个new,例如int** ary = new int*[row_num];for(int i=0: i< row num; ++1)ary [i] new int [col_num]用 vector的话一行代码搞定vector>ary(row_num, vector(col_num, 0))使用 reserve来避免不必要的重新分配第2章线性表这类题目考察线性表的操作,例如,数组,单链表,双向链表等。2数组2.1.1 Remove Duplicates from Sorted array描述Given a sorted array, remove the duplicates in place such that each element appear only onceand return the new lengthDo not allocate extra space for another array, you must do this in place with constant memoryFor example, Given input array A =[1, 1, 2Your function should return length =2, and a is now [1, 2]分析无代码1/ LeetCode, Remove Duplicates from Sorted Array//时间复杂度0(n),空间复杂度0(1)class Solution tublicint removeDuplicates(vector& nums)tif (nums empty o) return 0;int index =ofor (int i =1: i nums size: 1++ iif (nums [index] ! nums [i])nums [++index]= nums [i]return index 12.1数组代码2//LeetCode, Remove Duplicates from Sorted Array/使用STL,时间复杂度0(n),空间复杂度0(1)class Solution ipublicint removeDuplicates(vector& nums)treturn distance(nums begin(), unique(nums begin(), nums end ())代码3/ LeetCode, Remove Duplicates from Sorted Array/使用STL,时间复杂度0(n),空间复杂度0(1)lass Solution fublicint removeDuplicates(vector& nums)treturn distance(nums begin(, removeDuplicates(nums begin(, nums end(), nums begintemplateOutIt removeDuplicates(InIt first, InIt last, OutIt output)thile (first last)i*output++ = *firstfirst upper_bound(first, last, *firstreturn output相关题目Remove Duplicates from Sorted Array I,见§2.1.22.1.2 Remove Duplicates from Sorted Array II描述Follow up for"Remove Duplicates " What if duplicates are allowed at most twice?For example, Given sorted array a =[1, 1, 1, 2, 2, 3]Your function should return length=5, and A is now [1, 1, 2, 2, 3分析加一个变量记录一下元素出现的次数即可。这题因为是已经排序的数组,所以一个变量即可解决。如果是没有排序的数组,则需要引入一个 hashmap来记录出现次数。4第2章线性表代码1// Leet Code, Remove Duplicates from Sorted Array II//时间复杂度0(n),空间复杂度0(1)//qauthorhex108(https://github.com/hex108)class Solution tublicint removeDuplicates(vector& nums)tif (nums size(<= 2) return nums sizeint index =2for (int i=2: i nums size(: i++)ff (nums [i] ! nums lindex -2]nums [index++]= nums [i]return index;代码2下面是一个更简洁的版本。上面的代码略长,不过扩展性好一些,例如将 occur<2改为 occur3,就变成了允许重复最多3次。// Leet Code, Remove Duplicates from Sorted Array II7/@author虞航仲(http://weibo.com/u/1666779725)//时间复杂度0(n),空间复杂度0(1)lass Solution fpublicint removeDuplicates(vector& nums)tconst int n=nums sizeint index =0:for (int i=0:i 0&&i