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Copy path581.shortest-unsorted-continuous-subarray.java
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581.shortest-unsorted-continuous-subarray.java
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/*
* @lc app=leetcode id=581 lang=java
*
* [581] Shortest Unsorted Continuous Subarray
*
* https://leetcode.com/problems/shortest-unsorted-continuous-subarray/description/
*
* algorithms
* Easy (30.41%)
* Likes: 1646
* Dislikes: 74
* Total Accepted: 79.4K
* Total Submissions: 260.7K
* Testcase Example: '[2,6,4,8,10,9,15]'
*
* 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 .
*
*
*/
class Solution {
public int findUnsortedSubarray(int[] nums) {
if (nums == null || nums.length <= 1) {
return 0;
}
int start = -1, end = -2, n = nums.length;
int max = nums[0];
int min = nums[n - 1];
for (int i = 1; i < n; i++) {
max = Math.max(max, nums[i]);
min = Math.min(min, nums[n - 1 - i]);
// end 索引后面的都是有序的
if (nums[i] < max) {
end = i;
}
// start 索引前面的都是有序的
if (nums[n - 1 - i] > min) {
start = n - 1 - i;
}
}
return end - start + 1;
}
}