leetcode 581: Shortest Unsorted Continuous Subarray

最新推荐文章于 2018-12-19 16:57:01 发布

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最新推荐文章于 2018-12-19 16:57:01 发布

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本文介绍了一种算法，该算法通过先排序整个数组，然后对比原数组找到需要排序的最短连续子数组，以确保整个数组有序。示例中输入数组为[2,6,4,8,10,9,15]，输出长度为5，意味着需要对子数组[6,4,8,10,9]进行排序。

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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:

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

注意：先排序再找变化了的元素

	public int findUnsortedSubarray(int[] nums) {
		ArrayList l = new ArrayList();
		for (int i = 0; i < nums.length; i++)
			l.add(nums[i]);
		l.sort(null);
		int sum = 0;
		int po_start = -1;
		int po_end = -2;
		for (int i = 0; i < nums.length; i++)
			if (nums[i] != (int) l.get(i))
				if (po_start == -1)
					po_start = i;
				else
					po_end = i;
		return po_end - po_start + 1;
	}