题目地址:https://leetcode-cn.com/problems/max-difference-you-can-get-from-changing-an-integer/题目描述给你一个整数 num 。你可以对它进行如下步骤恰好 两次 :选择一个数字 x (0 <= x <= 9). 选择另一个数字 y (0 <= y <= 9) 。数字 y 可以等于 x。 将num 中所有出现 x 的数位都用 y...
The input array will only contain0and1. The length of input array is a positive integer and will not exceed 10,000 思路:count存储当前连续1的数量,max为最大连续1的数量,遍历数组,遇见0便取count与max的较大者,遇见1则将count加1。要注意最后末尾的1。
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LeetCode python 链表 结点 转载 网络安全专家 4月前 16阅读 隔两个数取两个数python # 隔两个数取两个数的方法及其应用 ## 引言 在日常编程中,我们经常需要从一组数据中筛选出满足特定规则的数据。其中一种常见的需求是隔两个数取两个数,也就是从一组数据中每隔两个数取出两个数。本文将介绍这种方法的具...
The naive solution is brute-force, which is O((mn)^2). In order to be more efficient, I tried something similar to Kadane's algorithm. The only difference is that here we have upper bound restriction K. First, How to find the max sum rectangle in 2D array? The naive way is O(N...
findTheDifference.java │ ├── groupAnagrams.java │ ├── groupShiftedStrings.java │ ├── islandPerimeter.java │ ├── loggerRateLimiter.java │ ├── maximumSizeSubarraySumEqualsK.java │ ├── minimumWindowSubstring.java │ ├── sparseMatrixMultiplication.java │ ├── strobogram...
The naive solution is brute-force, which is O((mn)^2). In order to be more efficient, I tried something similar to Kadane's algorithm. The only difference is that here we have upper bound restriction K. First, How to find the max sum rectangle in 2D array? The naive way is O(N...
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