581. Shortest Unsorted Continuous Subarray

weekly contest 32的第一题。这题我一开始就想着把每个子串都拿出来sort一遍再放回去，跟sort完的对比一下。但是复杂度太高(O(n²)), TLE了(但是用熟了System.arraycopy(nums, 0, temp, 0, nums.length);这个方法。。)。
如下：

Approach 1: BRUTE FORCE(TIME LIMIT EXCEEDED)

    public int findUnsortedSubarray(int[] nums) {
        if (nums == null || nums.length == 0) return 0;

        int sorted[] = new int[nums.length];
        System.arraycopy(nums, 0, sorted, 0, nums.length);

        Arrays.sort(sorted);
        if (Arrays.equals(sorted, nums)) return 0;

        int minLen = nums.length;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                int temp[] = new int[nums.length];
                System.arraycopy(nums, 0, temp, 0, nums.length);
                Arrays.sort(temp, i, j + 1);
                if (Arrays.equals(sorted, temp)) {
                    minLen = Math.min(j + 1 - i, minLen);
                    break;
                }
            }
        }
        return minLen;
    }

Approach 2: Compare the sorted parts

从首位开始跟sort过的array相比。Time : O(nlogn)。

    public int findUnsortedSubarray(int[] nums) {
        int [] temp = new int[nums.length];
        System.arraycopy(nums,0 , temp , 0 , nums.length);
        Arrays.sort(temp);
        int start = 0 , end = nums.length -1 ;
        while (start< nums.length && temp[start] == nums[start] ){
            start ++ ;
        }
        while (end> start && temp[end] == nums[end]){
            end -- ;
        }
        return end + 1 - start;
    }

还看到有一种O(n)的解法，现在脑子晕了不想看了。

ref:
https://discuss.leetcode.com/topic/89282/java-o-n-time-o-1-space
https://discuss.leetcode.com/category/741/shortest-unsorted-continuous-subarray
Java拷贝数组的几种方法：
http://www.cnblogs.com/jjdcxy/p/5870524.html