1330. 翻转子数组得到最大的数组值

贪心法：

1. 翻转中间，则使用（最小区间最大值-最大区间最小值）*2 + 基准为数组值，这里不一定为正收益

2.翻转头或者尾，则位基准重新计算翻转点

基准=sum(abs(num[i] - num[i - 1]))  (1<=i<nums.size())

class Solution {
public:
    int maxValueAfterReverse(vector<int>& nums) {
        if (nums.size() <= 1)
            return 0;
        int sum = 0;
        int a = INT_MAX;
        int b = INT_MIN;
        for (int i = 1; i < nums.size(); ++i) {
            sum += abs(nums[i] - nums[i - 1]);
            a = min(a, max(nums[i], nums[i - 1]));
            b = max(b, min(nums[i], nums[i - 1]));
        }
        int ans = max(sum, sum + 2 * (b - a));
        for (int i = 1; i < nums.size() - 1; i++) {
            ans = max(ans, sum - abs(nums[i] - nums[i + 1]) + abs(nums[0] - nums[i + 1]));
            ans = max(ans, sum - abs(nums[i - 1] - nums[i]) + abs(nums[i - 1] - nums.back()));
        }
        return ans;
    }
};