博客
关于我
LeetCode.581 Shortest Unsorted Continuous Subarray
阅读量:806 次
发布时间:2019-03-17

本文共 2372 字,大约阅读时间需要 7 分钟。

To solve this problem, we need to find the shortest continuous subarray such that sorting this subarray in ascending order would make the entire array sorted in ascending order as well.

Approach

The approach to solve this problem involves the following steps:

  • Sort the Array: First, we create a sorted version of the input array. This helps us identify the segments of the array that are out of order.

  • Identify Differences: We compare the original array with the sorted array to find the indices where they first differ (start of the unsorted segment) and where they last differ (end of the unsorted segment).

  • Determine the Subarray Length: The length of the shortest subarray that needs to be sorted is given by the range from the first differing index to the last differing index, inclusive.

  • Solution Code

    public class Solution {    public int findMinimumSubarrayLength(int[] nums) {        int n = nums.length;        int[] sorted = Arrays.copyOf(nums, n);        Arrays.sort(sorted);                int start = 0;        while (start < n && sorted[start] == nums[start]) {            start++;        }                if (start >= n) {            return 0;        }                int end = n - 1;        while (end >= 0 && sorted[end] == nums[end]) {            end--;        }                return end - start + 1;    }}

    Explanation

  • Sorting the Array: We create a sorted version of the input array to compare against the original array and identify the unsorted segments.

  • Finding the Start of the Subarray: By iterating through the original array, we find the first index where the value does not match the corresponding value in the sorted array. This index marks the beginning of the segment that needs to be sorted.

  • Finding the End of the Subarray: Similarly, by iterating from the end of the array, we find the last index where the value does not match the corresponding value in the sorted array. This index marks the end of the segment that needs to be sorted.

  • Calculating the Length: The length of the subarray is calculated as the difference between the end and start indices, plus one.

  • This approach ensures that we efficiently find the shortest subarray that, when sorted, will result in the entire array being sorted. The time complexity is dominated by the sorting step, making it (O(n \log n)), which is efficient for large arrays up to 10,000 elements.

    转载地址:http://ekjez.baihongyu.com/

    你可能感兴趣的文章
    Mysql学习总结(78)——MySQL各版本差异整理
    查看>>
    Mysql学习总结(79)——MySQL常用函数总结
    查看>>
    Mysql学习总结(7)——MySql索引原理与使用大全
    查看>>
    Mysql学习总结(80)——统计数据库的总记录数和库中各个表的数据量
    查看>>
    Mysql学习总结(81)——为什么MySQL不推荐使用uuid或者雪花id作为主键?
    查看>>
    Mysql学习总结(82)——MySQL逻辑删除与数据库唯一性约束如何解决?
    查看>>
    Mysql学习总结(83)——常用的几种分布式锁:ZK分布式锁、Redis分布式锁、数据库分布式锁、基于JDK的分布式锁方案对比总结
    查看>>
    Mysql学习总结(84)—— Mysql的主从复制延迟问题总结
    查看>>
    Mysql学习总结(85)——开发人员最应该明白的数据库设计原则
    查看>>
    Mysql学习总结(8)——MySql基本查询、连接查询、子查询、正则表达查询讲解
    查看>>
    Mysql学习总结(9)——MySql视图原理讲解与使用大全
    查看>>
    Mysql学习笔记 - 在Centos7环境下离线安装Mysql
    查看>>
    MySQL学习笔记十七:复制特性
    查看>>
    Mysql学习第一课-mysql的定义及sql语句
    查看>>
    mysql学号的字符长度_MYSQL--2
    查看>>
    mysql安全模式: sql_safe_updates
    查看>>
    mysql安装,卸载,连接
    查看>>
    MySQL安装之没有配置向导
    查看>>
    mysql安装出现 conflicts with mysql*的解决办法
    查看>>
    mysql安装卡在最后一步解决方案(附带万能安装方案)
    查看>>