Solution：
思路：
Time Complexity: O(N) Space Complexity: O(N)
Solution Code：
class Solution {
public int findUnsortedSubarray(int[] A) {
int n = A.length;
int begin = -1, end = -2;
int 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) begin = n - 1 - i;
}
return end - begin + 1;
}
}
