在[1,m]中的所有数，只需要考虑1和m的插入位置，其他数对结果不会产生影响。

#include <algorithm> #include <vector> class Solution { public:     long long min_value(vector<int>& nums, int m) {         int n = nums.size();         using ll = long long;         ll d = 0;         for (int i = 1; i < n; i++) {             d += abs(nums[i] - nums[i-1]);         }         ll delta1 = min(abs(nums[0] - m), abs(m - nums.back()));         ll delta2 = min(abs(nums[0] - 1), abs(1 - nums.back()));         for (int i = 1; i < n; i++) {             ll c1 = abs(nums[i] - nums[i-1]);             ll c2 = abs(m - nums[i-1]) + abs(nums[i] - m);             delta1 = min(delta1, c2 - c1);         }         for (int i = 1; i < n; i++) {             ll c1 = abs(nums[i] - nums[i-1]);             ll c2 = abs(1 - nums[i-1]) + abs(nums[i] - 1);             delta2 = min(delta2, c2 - c1);         }         return d + delta1 + delta2;     } };