You are given an integer array nums. The value of this array is defined as the sum of |nums[i] - nums[i + 1]| for all 0 <= i < nums.length - 1.
You are allowed to select any subarray of the given array and reverse it. You can perform this operation only once.
Find maximum possible value of the final array.
Input: nums = [2,3,1,5,4] Output: 10 Explanation: By reversing the subarray [3,1,5] the array becomes [2,5,1,3,4] whose value is 10.
Input: nums = [2,4,9,24,2,1,10] Output: 68
1 <= nums.length <= 3 * 104-105 <= nums[i] <= 105
implSolution{pubfnmax_value_after_reverse(nums:Vec<i32>) -> i32{let origin = (1..nums.len()).map(|i| (nums[i] - nums[i - 1]).abs()).sum::<i32>();letmut max_num = nums[0].min(nums[1]);letmut min_num = nums[0].max(nums[1]);letmut ret = origin;for i in1..nums.len() - 1{let a = nums[i - 1];let b = nums[i];let c = nums[i + 1]; ret = ret.max(origin + (c - nums[0]).abs() - (c - b).abs()); ret = ret.max(origin + (a - nums.last().unwrap()).abs() - (b - a).abs());let x = b.max(c);let y = b.min(c);if x < max_num { ret = ret.max(origin + 2* max_num - 2* x);}if y > min_num { ret = ret.max(origin + 2* y - 2* min_num);} max_num = max_num.max(y); min_num = min_num.min(x);} ret }}