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数据库安装配置与常用命令
    查看>>
    MySQL数据库实现主从同步数据
    查看>>
    mysql数据库导入导出_windows系统以及linux系统下的操作---linux工作笔记042
    查看>>
    mysql数据库导出导入
    查看>>
    MySQL数据库工具类之——DataTable批量加入MySQL数据库(Net版)
    查看>>
    mysql数据库常用命令
    查看>>
    MySQL数据库必会的增删查改操作(CRUD)
    查看>>
    MySQL数据库性能分析与调优实践
    查看>>
    mysql数据库扫盲,你真的知道什么是数据库嘛
    查看>>
    mysql数据库批量插入数据shell脚本实现
    查看>>
    MySQL数据库操作
    查看>>
    MySQL数据库故障排错
    查看>>
    MySQL数据库无法远程连接的解决办法
    查看>>
    mysql数据库时间类型datetime、bigint、timestamp的查询效率比较
    查看>>
    MySQL数据库服务器端核心参数详解和推荐配置(一)
    查看>>
    mysql数据库死锁的产生原因及解决办法
    查看>>
    MySQL数据库的事务管理
    查看>>
    mysql数据库的备份与恢复
    查看>>
    Mysql数据库的条件查询语句
    查看>>
    MySQL数据库的高可用
    查看>>