LeetCode每日一题(1014. Best Sightseeing Pair)

You are given an integer array values where values[i] represents the value of the ith sightseeing spot. Two sightseeing spots i and j have a distance j - i between them.

The score of a pair (i < j) of sightseeing spots is values[i] + values[j] + i - j: the sum of the values of the sightseeing spots, minus the distance between them.

Return the maximum score of a pair of sightseeing spots.

Example 1:

Input: values = [8,1,5,2,6]
Output: 11

Explanation: i = 0, j = 2, values[i] + values[j] + i - j = 8 + 5 + 0 - 2 = 11

Example 2:

Input: values = [1,2]
Output: 2

Constraints:

• 2 <= values.length <= 5 * 104
• 1 <= values[i] <= 1000

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从右往左遍历， 一旦发现最大值， 伴随着每一步，最大值-1， 因为一旦有新的最大值加入进来， 旧的最大值就永远都不会再变为最大值， 所以我们只需要维护一个变量来保存当前的最大值即可

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impl Solution {
    pub fn max_score_sightseeing_pair(values: Vec<i32>) -> i32 {
        let mut ans = 0;
        let mut max = 0;
        for v in values.into_iter().rev() {
            if max != 0 {
                max -= 1;
                ans = ans.max(v + max);
                max = max.max(v);
                continue;
            }
            max = max.max(v);
        }
        ans
    }
}