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
    mysqldump实现数据备份及灾难恢复
    查看>>
    mysqldump数据库备份无法进行操作只能查询 --single-transaction
    查看>>
    mysqldump的一些用法
    查看>>
    mysqli
    查看>>
    MySQLIntegrityConstraintViolationException异常处理
    查看>>
    mysqlreport分析工具详解
    查看>>
    MySQLSyntaxErrorException: Unknown error 1146和SQLSyntaxErrorException: Unknown error 1146
    查看>>
    Mysql_Postgresql中_geometry数据操作_st_astext_GeomFromEWKT函数_在java中转换geometry的16进制数据---PostgreSQL工作笔记007
    查看>>
    mysql_real_connect 参数注意
    查看>>
    mysql_secure_installation初始化数据库报Access denied
    查看>>
    MySQL_西安11月销售昨日未上架的产品_20161212
    查看>>
    Mysql——深入浅出InnoDB底层原理
    查看>>
    MySQL“被动”性能优化汇总
    查看>>
    MySQL、HBase 和 Elasticsearch:特点与区别详解
    查看>>
    MySQL、Redis高频面试题汇总
    查看>>
    MYSQL、SQL Server、Oracle数据库排序空值null问题及其解决办法
    查看>>
    mysql一个字段为空时使用另一个字段排序
    查看>>
    MySQL一个表A中多个字段关联了表B的ID,如何关联查询?
    查看>>
    MYSQL一直显示正在启动
    查看>>
    MySQL一站到底!华为首发MySQL进阶宝典,基础+优化+源码+架构+实战五飞
    查看>>