Leetcode

835. Image Overlap Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: Array、Matrix 一、題目 You are given two images, img1 and img2, represented as binary, square matrices of size n x n. A binary matrix has only 0s and 1s as values. We translate one image however we choose by sliding all the 1 bits left, right, up, and/or down any number of units. We then place it on top of the other image. We can then calculate the overlap by counting the number of positions that have a 1 in both images. Note also that a translation does not include any kind of rotation. Any 1 bits that are translated outside of the matrix bordered are erased. Return the largest possible overlap. ...

12. Integer to Roman Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: Hash Table、Math、String 一、題目 Roman numerals are represented by seven different symbols: I, V, X, L, C, D, and M. \(\boxed{\begin{array}{ll} \textbf{Symbol}&\textbf{Value}\\ \texttt{I}&1\\ \texttt{V}&5\\ \texttt{X}&10\\ \texttt{L}&50\\ \texttt{C}&100\\ \texttt{D}&500\\ \texttt{M}&1000\\ \end{array}}\) For example, 2 is written as II in Roman numeral, just two one’s added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II. Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used: ...

11. Container With Most Water Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: Array、Two Pointer、Greedy 一、題目 You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]). Find two lines that together with the x-axis form a container, such that the container contains the most water. Return the maximum amount of water a container can store. Notive that you may not slant the container. ...

10. Regular Expression Matching Hardness: \(\color{red}\textsf{Hard}\) Ralated Topics: String、Dynamic Programming、Recursion 一、題目 Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where: '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). Example 1: Input: s = “aa”, p = “a” Output: false Explanation: “a” does not match the entire string “aa”. Example 2: ...

9. Palindrome Number Hardness: \(\color{green}\textsf{Easy}\) Ralated Topics: Math 一、題目 Given an integer x, return true if x is palindrome number. An integer is a palindrome when it reads the same backward as forward. For example, 121 is a palindrome while 123 is not. Example 1: Input: x = 121 Output: true Explanation: 121 reads as 121 from left to right and from right to left. Example 2: Input: x = -121 Output: false Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome. Example 3: ...

8. String to Integer (atoi) Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: String 一、題目 Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++’s atoi function). The algorithm for myAtoi(string s) is as follows: Read in and ignore any leading whitespace. Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored. Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2). If the integer is out of the 32-bit signed integer range [-2^31, 2^31-1], then clamp the integer so that it remains in the range. Specifically, integers less than -2^31 should be clamped to -2^31, and integer greater than 2^31-1 should be clamped to 2^31-1. Return the integer as the final result. Note: ...

7. Reverse Integer Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: Math 一、題目 Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-2^31, 2^31-1], then return 0. Assume the environment does not allow you to store 64-bit integers (signed or unsigned). Example 1: Input: x = 123 Output: 321 Example 2: Input: x = -123 Output: -321 Example 3: ...

6. Zigzag Conversion Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: String 一、題目 The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to dispaly this pattern in a fixed font for better legibility) \( \quad\texttt{P A H N}\\ \quad\texttt{APLSIIG}\\ \quad\texttt{Y I R}\\ \) And then read line by line: "PAHNAPLSIIGYIR" Write the code that will take a string and make this conversion given a number of rows: string convert(string s, int numRows ...

5. Longest Substring Without Repeating Characters Hardness: \(\color{orange}\textsf{Medium}\) Ralated Topics: String、Dynamic Programming 一、題目 Given a string s, return the longest palindromic substring in s. A string is called a palindrome string if the reverse of that string is the same of the original string. Example 1: Input: s = “babad” Output: “bab” Explanation: “aba” is also a valid answer. Example 2: Input: s = “cbbd” Output: “bb” Constraints: 1 <= s.length <= 1000 s consists of only digits and English letters. 二、分析 注意 palindrome string 的特性： 當長度為 1 時，必為 palindrome string 當長度為 2 時，兩個字元必須相同才為 palindrome string 當長度 >2 時，palindrome string 必須滿足 最左邊的字元等於最右邊的字元，即 s[left] == s[right] 除去最左邊的字元跟最右邊的字元，必須為 palindrome string， 即 s.substr(left+1, len-2) 為 palindromic。 三、解題 1. Dynamic Prograimming Time complexity: \(O(n^2)\) Space complexity: \(O(n^2)\) string longestPalindrome(string s) { int n = s.length(); string res; bool dp[n][n]; memset(dp, false, sizeof(dp)); int len = 0; for (int j = 0; j < n; j++) { for (int i = 0; i <= j; i++) { if (i == j) { // 長度為 1 dp[i][j] = true; } else if (j - i == 1) { // 長度為 2 dp[i][j] = s[i] == s[j]; } else { // 長度 > 2 dp[i][j] = s[i] == s[j] && dp[i+1][j-1]; } if (dp[i][j] && j - i + 1 > len) { // 比較長度 len = j - i + 1; res = s.substr(i, len); } } } return res; } 回目錄 Catalog ...

4. Median of Two Sorted Arrays Hardness: \(\color{red}\textsf{Hard}\) Ralated Topics: Array、Binary Search、Divide and Conquer 一、題目 Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: Input: nums1 = [1,3], nums2 = [2] Output: 2.00000 Explanation: merged array = [1,2,3] and median is 2. Example 2: Input: nums1 = [1,2], nums2 = [3,4] Output: 2.50000 Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5. Constraints: ...