581. Shortest Unsorted Continuous Subarray

Given an integer array, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order, too.

You need to find the shortest such subarray and output its length.

Example 1:

Input: [2, 6, 4, 8, 10, 9, 15]
Output: 5
Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.

Note:

1. Then length of the input array is in range [1, 10,000].
2. The input array may contain duplicates, so ascending order here means <=.

Approach #1: Math. [Java]

public int findUnsortedSubarray(int[] A) {
    int n = A.length, beg = -1, end = -2, min = A[n-1], max = A[0];
    for (int i=1;i<n;i++) {
      max = Math.max(max, A[i]);
      min = Math.min(min, A[n-1-i]);
      if (A[i] < max) end = i;
      if (A[n-1-i] > min) beg = n-1-i; 
    }
    return end - beg + 1;
}

Analysis:

So brilliant.

Using the variables beg and end to keep track the min subarray A[beg .... end] which must be sort for the entire array A to be sorted, if end < beg < 0 at the end in the for loop. Then the array is already full sorted.

Reference:

https://leetcode.com/problems/shortest-unsorted-continuous-subarray/discuss/103057/Java-O(n)-Time-O(1)-Space

标签：end,Continuous,beg,ascending,Unsorted,subarray,Subarray,array,order
来源： https://www.cnblogs.com/ruruozhenhao/p/10645919.html