LeetCode刷题【Array】 Shortest Unsorted Continuous Subarray

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本文介绍了解决LeetCode上一道关于寻找最短无序子数组问题的两种方法。该问题要求找出一个整数数组中只需排序即可使整个数组有序的最短连续子数组长度。文中提供了两个Java实现方案，分别采用不同的逻辑来解决问题。

【题目】

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 <=.

方法一：

public class Solution {
    public int findUnsortedSubarray(int[] nums) {
        Stack < Integer > stack = new Stack < Integer > ();
        int l = nums.length, r = 0;
        for (int i = 0; i < nums.length; i++) {
            while (!stack.isEmpty() && nums[stack.peek()] > nums[i])
                l = Math.min(l, stack.pop());
            stack.push(i);
        }
        stack.clear();
        for (int i = nums.length - 1; i >= 0; i--) {
            while (!stack.isEmpty() && nums[stack.peek()] < nums[i])
                r = Math.max(r, stack.pop());
            stack.push(i);
        }
        return r - l > 0 ? r - l + 1 : 0;
    }
}

方法二：

public int findUnsortedSubarray(int[] A) {
    int n = A.length, beg = -1, end = -2, min = A[n-1], max = A[0];
    for (int i=1;i<n;i++) {
      max = Math.max(max, A[i]);
      min = Math.min(min, A[n-1-i]);
      if (A[i] < max) end = i;
      if (A[n-1-i] > min) beg = n-1-i; 
    }
    return end - beg + 1;
}

参考 compton_scatter ；解答

【参考】

https://leetcode.com/problems/can-place-flowers/#/description