LeetCode - 581. Shortest Unsorted Continuous Subarray

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 <=.

找到第一個小於自身左邊的下標，第一個大於自身右邊的下標，求差。

class Solution {
    public int findUnsortedSubarray(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        int len = nums.length, start = 1, end = 0;
        for (int i=0; i<len; i++)
            if (!judge(nums, i)) {
                start = i;
                break;
            }
        for (int i=len-1; i>start; i--)
            if (!judge(nums, i)) {
                end = i;
                break;
            }
        return end - start + 1;
    }

    private boolean judge(int[] nums, int index) {
        for (int i=0; i<index; i++)
            if (nums[i] > nums[index])
                return false;
        for (int i=index+1; i<nums.length; i++)
            if (nums[i] < nums[index])
                return false;
        return true;
    }
}

簡潔優化版：

class Solution {
    public int findUnsortedSubarray(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        int len = nums.length, start = -1, end = -2;
        int min = nums[len-1], max = nums[0];
        for (int i=0; i<len; i++) {
            max = Math.max(nums[i], max);
            min = Math.min(nums[len - i - 1], min);
            if (nums[i] < max)
                end = i;
            if (nums[len - i - 1] > min)
                start = len - i - 1;
        }
        return end - start + 1;
    }

}