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 分页语句 Limit原理
    查看>>
    MySQL 创建新用户及授予权限的完整流程
    查看>>
    mysql 创建表,不能包含关键字values 以及 表id自增问题
    查看>>
    mysql 删除日志文件详解
    查看>>
    mysql 判断表字段是否存在,然后修改
    查看>>
    mysql 协议的退出命令包及解析
    查看>>
    mysql 取表中分组之后最新一条数据 分组最新数据 分组取最新数据 分组数据 获取每个分类的最新数据
    查看>>
    mysql 多个表关联查询查询时间长的问题
    查看>>
    mySQL 多个表求多个count
    查看>>
    mysql 多字段删除重复数据,保留最小id数据
    查看>>
    MySQL 多表联合查询:UNION 和 JOIN 分析
    查看>>
    MySQL 大数据量快速插入方法和语句优化
    查看>>
    mysql 如何给SQL添加索引
    查看>>
    mysql 字段区分大小写
    查看>>
    mysql 字段合并问题(group_concat)
    查看>>
    mysql 字段类型类型
    查看>>
    MySQL 字符串截取函数,字段截取,字符串截取
    查看>>
    MySQL 存储引擎
    查看>>
    mysql 存储过程 注入_mysql 视图 事务 存储过程 SQL注入
    查看>>
    MySQL 存储过程参数:in、out、inout
    查看>>