581. Shortest Unsorted Continuous Subarray

寻找最短的未排序数组。找到未排序数组不难，但是如何寻找最短的？

Approach 1 : Sort

首先对数组进行排序，然后将未排序数组与原数组进行比较
Runtime: 32 ms, faster than 67.07% of C++ online submissions

class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
        vector<int> sorted(nums);
        sort(sorted.begin(), sorted.end());
        int n = nums.size(), i = 0, j = n - 1;
        while (i < n && nums[i] == sorted[i])
            i++;
        while (j > i && nums[j] == sorted[j])
            j--;
        return j + 1 - i;
    }
};

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Approach 2 : max on left, min on right

Runtime: 24 ms, faster than 98.78% of C++ online submissions

The idea is that:

1. From the left, for a number to stay untouched, there need to be no number less than it on the right hand side;
2. From the right, for a number to stay untouched, there need to be no number greater than it on the left hand side;
   Those 2 can be easily checked if we build 2 vectors:
3. max so far from the left,
4. and min so far from the right;

/**
 *            /------------\
 * nums:  [2, 6, 4, 8, 10, 9, 15]
 * minr:   2  4  4  8   9  9  15
 *         <--------------------
 * maxl:   2  6  6  8  10 10  15
 *         -------------------->
 */
class Solution {
public:
    int findUnsortedSubarray(vector<int>& nums) {
        int n = nums.size();
        vector<int> maxlhs(n);   // max number from left to cur
        vector<int> minrhs(n);   // min number from right to cur
        for (int i = n - 1, minr = INT_MAX; i >= 0; i--) minrhs[i] = minr = min(minr, nums[i]);
        for (int i = 0,     maxl = INT_MIN; i < n;  i++) maxlhs[i] = maxl = max(maxl, nums[i]);

        int i = 0, j = n - 1;
        while (i < n && nums[i] <= minrhs[i]) i++;
        while (j > i && nums[j] >= maxlhs[j]) j--;

        return j + 1 - i;
    }
};