581. Shortest Unsorted Continuous Subarray 

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class Solution {
    public int findUnsortedSubarray(int[] nums) {
        // [2, 6, 4, 8, 10, 9, 15]
        //    [6, 4, 8, 10, 9]

        // [1, 3, 2, 2, 2]
        //    [2, 2, 2, 3]

        // [1, 2, 3]
        // [1, 2, 3]

        int[] sortedNums = nums.clone();
        Arrays.sort(sortedNums);

        int i = 0, j = nums.length - 1;
        while (i <= j && nums[i] == sortedNums[i]) {
            i++;
        }
        while (i <= j && nums[j] == sortedNums[j]) {
            j--;
        }

        return j - i + 1;
    }
}
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class Solution {
    public int findUnsortedSubarray(int[] nums) {
        // [2, 16, 4, 0, 10, 9, 15]
        // [2, 16, 4, 0, 10, 9, 15]

        // [1, 3, 2, 2, 2]
        //    [2, 2, 2, 3]

        // [1, 2, 3]
        // [1, 2, 3]

        int left = 0, right = -1;
        int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;

        for (int i = 0; i < nums.length; i++) {
            if (nums[i] < max) {
                // nums[i] 比左侧的最大值小，nums[i] 必须纳入子数组
                right = i;
            } else {
                max = nums[i];
            }
            if (nums[nums.length - 1 - i] > min) {
                // nums[nums.length - 1 - i] 比右侧的最小值大，nums[nums.length - 1 - i] 必须纳入子数组
                left = nums.length - 1 - i;
            } else {
                min = nums[nums.length - 1 - i];
            }
        }

        return right - left + 1;
    }
}

Reference

581. Shortest Unsorted Continuous Subarray