LeetCode 581. Shortest Unsorted Contimuous Subarray

class Solution {public:   int findUnsortedSubarray(vector<int>& nums){vector<int> num_buffer;for (int i = 0; i < nums.size(); i++){num_buffer.push_back(nums[i]);}int temp_i;for (int i = 0; i < nums.size(); i++){for (int j = i+1; j < nums.size(); j++){if (num_buffer[j] < num_buffer[i]){temp_i = num_buffer[j];num_buffer[j] = num_buffer[i];num_buffer[i] = temp_i;}}}int begin = 0;int end = 0;bool flag = true;for (int i = 0; i < nums.size(); i++){if (nums[i] != num_buffer[i] && flag){begin = i;flag = false;}else if (nums[i] != num_buffer[i]){end = i;}}if (begin != end)return (end - begin + 1);elsereturn 0;}};

思路：先把数组进行排序，然后把排序完的数组和排序之前的数组进行比较。

solution上提供的一个解法，时间复杂度确实比我的解法小。

int findUnsortedSubarray(vector<int>& nums) {        int shortest = 0;                int left = 0, right = nums.size() - 1;        while (left < nums.size() - 1 && nums[left] <= nums[left + 1]) { left++; }        while (right > 0 && nums[right] >= nums[right - 1]) { right--; };                if (right > left) {            int vmin = INT_MAX, vmax = INT_MIN;            for (int i = left; i <= right; ++i) {                if (nums[i] > vmax) {                    vmax = nums[i];                }                if (nums[i] < vmin) {                    vmin = nums[i];                }            }                        while (left >= 0 && nums[left] > vmin) { left--; };            while (right < nums.size() && nums[right] < vmax) { right++; };                        shortest = right - left - 1;        }                return shortest;    }