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