1330.翻转子数组得到最大的数组值

链接：1330.翻转子数组得到最大的数组值
难度：Hard
标签：贪心、数组、数学
简介：给你一个整数数组 nums 。「数组值」定义为所有满足 0 <= i < nums.length-1 的 |nums[i]-nums[i+1]| 的和。

题解 1 - cpp

• 编辑时间：2023-05-12
• 执行用时：84ms
• 内存消耗：39.3MB
• 编程语言：cpp
• 解法介绍：https://leetcode.cn/problems/reverse-subarray-to-maximize-array-value/solution/bu-hui-hua-jian-qing-kan-zhe-pythonjavac-c2s6/。

class Solution {
public:
    int maxValueAfterReverse(vector<int>& nums) {
        int n = nums.size(), sums = 0, nmax = INT_MIN, nmin = INT_MAX, val = 0;
        for (int i = 1; i < n; i++) {
            int num = abs(nums[i] - nums[i - 1]);
            sums += num;
            nmax = max(nmax, min(nums[i], nums[i - 1]));
            nmin = min(nmin, max(nums[i], nums[i - 1]));
            val = max(val, max(abs(nums[i] - nums[0]), abs(nums[i - 1] - nums[n - 1])) - num);
        }
        return sums + max(val, 2 * (nmax - nmin));
    }
};

题解 2 - python

• 编辑时间：2023-05-12
• 执行用时：328ms
• 内存消耗：19.6MB
• 编程语言：python
• 解法介绍：同上。

class Solution:
def maxValueAfterReverse(self, nums: List[int]) -> int:
    n = len(nums)
    sums = 0
    nmax = -inf
    nmin = inf
    val = 0
    for i in range(1, n):
        num = abs(nums[i] - nums[i - 1])
        sums += num
        nmax = max(nmax, min(nums[i], nums[i - 1]))
        nmin = min(nmin, max(nums[i], nums[i - 1]))
        val = max(val, max(abs(nums[i] - nums[0]), abs(nums[i - 1] - nums[n - 1])) - num)
    return sums + max(val, 2 * (nmax - nmin))

题解 3 - rust

• 编辑时间：2023-05-12
• 执行用时：8ms
• 内存消耗：2.3MB
• 编程语言：rust
• 解法介绍：同上。

impl Solution {
    pub fn max_value_after_reverse(nums: Vec<i32>) -> i32 {
        use std::cmp::{max, min};
        let n = nums.len();
        let mut sums = 0;
        let mut nmax = i32::MIN;
        let mut nmin = i32::MAX;
        let mut val = 0;
        for i in 1..n {
            let num = (nums[i] - nums[i - 1]).abs();
            sums += num;
            nmax = max(nmax, min(nums[i], nums[i - 1]));
            nmin = min(nmin, max(nums[i], nums[i - 1]));
            val = max(
                val,
                max((nums[i] - nums[0]).abs(), (nums[i - 1] - nums[n - 1]).abs()) - num,
            );
        }
        sums + max(val, 2 * (nmax - nmin))
    }
}