[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] Two arrays are called similar if one can be obtained from another by swapping at most one pair of elements in one of the arrays. Given two arrays a and b, check whether they are similar. Example: For a = [1,...
[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] You are given an array of integers. On each move you are allowed to increase exactly one of its element by one. Find the minimal number of moves required to obtain a strictly increasing sequence from the input. Example: For...
[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] 题目如下： Given an array of strings, return another array containing all of its longest strings. Example: For inputArray = ["aba", "aa", "ad", "vcd", "aba"], the output should be allLongestStrings(inputArray) = ["aba", "vcd", "aba"]. Input/Output: [execution time limit] 4 seconds (js) [input] array.string inputArray A non-empty...
[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] Objective Given two strings, find the number of common characters between them. Example For s1 = "aabcc" and s2 = "adcaa", the output should be commonCharacterCount(s1, s2) = 3. Strings have 3 common characters – 2 “a”s and 1 “c”. Input/Output [execution time limit] 4 seconds (py3) [input] string s1A string consisting of lowercase English...
[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] 题目如下： Ticket numbers usually consist of an even number of digits. A ticket number is considered lucky if the sum of the first half of the digits is equal to the sum of the second half. Given a ticket number...
[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] Given the head of a linked list, rotate the list to the right by k places. Example 1: Input: head = [1,2,3,4,5], k = 2 Output: [4,5,1,2,3] Example 2: Input: head = [0,1,2], k = 4 Output: [2,0,1] Constraints: The number of nodes in the list is in the range [0, 500]. -100...
[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] Given a positive integer n, generate an n x n matrix filled with elements from 1 to n2 in spiral order. Example 1: Input: n = 3 Output: [[1,2,3],[8,9,4],[7,6,5]] Example 2: Input: n = 1 Output: [[1]] Constraints: 1 <= n <= 20 解题思路 根据方向和下一个填充的位置的对应，找到下一个位置，如果这个位置出边界或已经被填过，则转换方向 代码 class Solution: def generateMatrix(self, n: int) -> List[List[int]]: ...
[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] Given an m x n matrix, return all elements of the matrix in spiral order. Example 1: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,3,6,9,8,7,4,5] Example 2: Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]] Output: [1,2,3,4,8,12,11,10,9,5,6,7] Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 10 -100 <= matrix[i][j] <= 100 题目大意顺时针螺旋着打印二维矩阵。 解题方法维护四个边界和运动方向螺旋打印，一定会在遍历的时候更改方向。在什么时候更改方向呢？在最外圈运动的时候是到达边界的时候。但是当移动到Example 1中4的位置时，要向右移动（而不是向上），那么相当于上边界已经移动了第二行。 同理，我们推断： 我们维护四个边界left, right, up, down，表示尚未走过的、可以移动的矩阵范围，起始时四个边界即矩阵的边界。当每次遇到新的边界的时候，需要把移动方向顺时针旋转90度，同时把刚刚走过的那个边界线（这条边界线上所有元素已经遍历过）需要向矩阵内移动，即缩小了边界。当所有的位置都被遍历了一次，则停止。 python代码如下，核心是每次遇到新的边界时，顺时针修改移动方向，并且将老边界内移。 class Solution(object):def...
[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] Implement pow(x, n), which calculates x raised to the power n (i.e., xn). Example 1: Input: x = 2.00000, n = 10 Output: 1024.00000 Example 2: Input: x = 2.10000, n = 3 Output: 9.26100 Example 3: Input: x = 2.00000, n = -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25 Constraints: -100.0...
[vc_row css_animation="" row_type="row" use_row_as_full_screen_section="no" type="full_width" angled_section="no" text_align="left" background_image_as_pattern="without_pattern"][vc_column][vc_column_text] Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target. Each number in candidates may only be used once in the combination. Note: The solution set must not contain duplicate combinations. Example 1: Input: candidates...