LeetCode - 581. Shortest Unsorted Continuous Subarray

Posted 2021-01-13 码上哈希

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 <=.

找到第一个小于自身左边的下标，第一个大于自身右边的下标，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;
    }
}

简洁优化版：

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;
    }

}