README_CN.md

1793. 好子数组的最大分数

给你一个整数数组 nums **（下标从 0 开始）**和一个整数 k 。

一个子数组 (i, j) 的 分数 定义为 min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1) 。一个 好 子数组的两个端点下标需要满足 i <= k <= j 。

请你返回 好 子数组的最大可能 分数 。

示例 1:

输入: nums = [1,4,3,7,4,5], k = 3
输出: 15
解释: 最优子数组的左右端点下标是 (1, 5) ，分数为 min(4,3,7,4,5) * (5-1+1) = 3 * 5 = 15 。

示例 2:

输入: nums = [5,5,4,5,4,1,1,1], k = 0
输出: 20
解释: 最优子数组的左右端点下标是 (0, 4) ，分数为 min(5,5,4,5,4) * (4-0+1) = 4 * 5 = 20 。

提示:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 2 * 104
• 0 <= k < nums.length

题解 (Rust)

1. 题解

impl Solution {
    pub fn maximum_score(nums: Vec<i32>, k: i32) -> i32 {
        let mut i = k;
        let mut j = k;
        let mut min_num = nums[k as usize];
        let mut ret = min_num;

        loop {
            while i - 1 >= 0 && nums[i as usize - 1] >= min_num {
                i -= 1;
            }
            while j + 1 < nums.len() as i32 && nums[j as usize + 1] >= min_num {
                j += 1;
            }

            ret = ret.max(min_num * (j - i + 1));
            i -= 1;
            if i < 0 {
                break;
            }
            min_num = nums[i as usize];
        }

        i = k;
        j = k;
        min_num = nums[k as usize];

        loop {
            while i - 1 >= 0 && nums[i as usize - 1] >= min_num {
                i -= 1;
            }
            while j + 1 < nums.len() as i32 && nums[j as usize + 1] >= min_num {
                j += 1;
            }

            ret = ret.max(min_num * (j - i + 1));
            j += 1;
            if j >= nums.len() as i32 {
                break;
            }
            min_num = nums[j as usize];
        }

        ret
    }
}