LeetCode581. Shortest Unsorted Continuous Subarray

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Description

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 <=.

my program

思路：构建一个排序好的数组，然后与原数组进行对比，找出最先和最后不同的元素，相减+1即为所求答案。此算法时间复杂度是O(nlogn)，空间复杂度是O(n).

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

Submission Details
307 / 307 test cases passed.
Status: Accepted
Runtime: 56 ms

解法二

/**
 *            /------------
 * 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;
    }
};

此算法时间复杂度仅是O(n)，空间复杂度是O(n). 优于第一种算法