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
    mutiplemap 总结
    查看>>
    MySQL Error Handling in Stored Procedures---转载
    查看>>
    MVC 区域功能
    查看>>
    MySQL FEDERATED 提示
    查看>>
    mysql generic安装_MySQL 5.6 Generic Binary安装与配置_MySQL
    查看>>
    Mysql group by
    查看>>
    MySQL I 有福啦,窗口函数大大提高了取数的效率!
    查看>>
    mysql id自动增长 初始值 Mysql重置auto_increment初始值
    查看>>
    MySQL in 太多过慢的 3 种解决方案
    查看>>
    Mysql Innodb 锁机制
    查看>>
    MySQL InnoDB中意向锁的作用及原理探
    查看>>
    MySQL InnoDB事务隔离级别与锁机制深入解析
    查看>>
    Mysql InnoDB存储引擎 —— 数据页
    查看>>
    Mysql InnoDB存储引擎中的checkpoint技术
    查看>>
    Mysql InnoDB存储引擎中缓冲池Buffer Pool、Redo Log、Bin Log、Undo Log、Channge Buffer
    查看>>
    MySQL InnoDB引擎的锁机制详解
    查看>>
    Mysql INNODB引擎行锁的3种算法 Record Lock Next-Key Lock Grap Lock
    查看>>
    mysql InnoDB数据存储引擎 的B+树索引原理
    查看>>
    mysql interval显示条件值_MySQL INTERVAL关键字可以使用哪些不同的单位值?
    查看>>
    mysql problems
    查看>>