[Leetcode] 581. Shortest Unsorted Continuous Subarray 解题报告

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/78922500

本文介绍了一种高效算法，用于找出一个整数数组中最短的连续子数组，仅对该子数组排序即可使整个数组有序。算法通过三个步骤实现：定位首个非增位置、确定子数组的最大最小值并进行扩展。

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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 <=.

思路：

分三步骤：1）分别从左和从右，找出第一个非增/非降的位置left和right；2）在区间[left, right]之内找到最大值lmax和最小值rmin；3）将区间[left, right]分别向左和向右扩展，直到其左边的第一个数小于rmin，其右边的第一个数大于lmax。扩展后的区间[left, right]的长度就是题目所求。算法的时间复杂度是O(n)，空间复杂度是O(1)。

代码：

class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
        if (nums.size() == 0) {
            return 0;
        }
        int n = nums.size(), left = 0, right = n - 1;
        while (left < n - 1 && nums[left] <= nums[left + 1]) {
            ++left;
        }
        if (left == n - 1) {                        // already sorted
            return 0;
        }
        // We know the array is unsorted, so it's no need to judge right > 0
        while (nums[right] >= nums[right - 1]) {
            --right;
        }
        int rmin = INT_MAX, lmax = INT_MIN;         // the max/min value in [left, right]
        for (int i = left; i <= right; ++i) {
            if (nums[i] > lmax) {
                lmax = nums[i];
            }
            if (nums[i] < rmin) {
                rmin = nums[i];
            }
        }
        while (left >= 0 && nums[left] > rmin) {    // extends the left side
            --left;
        }
        while (right < n && nums[right] < lmax) {   // extends the right side
            ++right;
        }
        return right - left - 1;
    }
};