581. Shortest Unsorted Continuous Subarray

#### QUESTION:

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 <=.

#### EXPLANATION:

• 第一个解法

第一个解法比较简单，其实就是先排序之后再看一下不相同的位置。

• 第二个解法

1.由前向后决定end的位置，因为最后的位置肯定应该是最大值，如果不是的话，那么就确定end就是当前位置。

2.右后向前决定start的位置，与上面相反，就可以确定出start的位置，最后进行处理就可以了。

#### SOLUTION:

``````public class Solution {
// SOLUTION 1:
public static int findUnsortedSubarray(int[] nums) {
if(nums.length == 0|| nums.length ==1) return 0;
int[] copy = Arrays.copyOf(nums,nums.length);
Arrays.sort(copy);
int m = -1,n = -2;
for(int i = 0;i<nums.length;i++){
if(copy[i]!=nums[i]){
if(m == -1)
m = i;
n = i;
}
}
return n-m+1;
}
//SOLUTION 2:
public int findUnsortedSubarray(int[] nums) {
if(nums.length == 0|| nums.length ==1) return 0;

int n = nums.length,start = -1,end = -2,max = nums[0],min = nums[n-1];
for(int i = 1;i<nums.length;i++){
max = Math.max(max,nums[i]);
min = Math.min(min,nums[n-i-1]);
if(nums[i]<max) end = i;
if(min<nums[n-i-1]) start = n-i-1;
}
return end-start+1;
}
}
``````
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