[LeetCode] 1014. Best Sightseeing Pair
grandyang opened this issue · comments
Given an array A
of positive integers, A[i]
represents the value of the i
-th sightseeing spot, and two sightseeing spots i
and j
have distance j - i
between them.
The score of a pair (i < j
) of sightseeing spots is (A[i] + A[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: [8,1,5,2,6]
Output: 11
Explanation: i = 0, j = 2, `A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11`
Note:
2 <= A.length <= 50000
1 <= A[i] <= 1000
这道题给了一个正整数的数组A,定义了一种两个数字对儿的记分方式,为 A[i] + A[j] + i - j
,现在让找出最大的那组的分数。博主首先尝试了一下暴力搜索法,遍历所有的两个数的组合,计算得分,毫不意外的超时了,虽然只是一道 Medium 的题,仍然要给予 OJ 足够的尊重。那么来考虑如何优化一下,利用加法的分配律,可以得到 A[i] + i + A[j] - j
,为了使这个表达式最大化,A[i] + i
自然是越大越好,这里可以使用一个变量 mx 来记录之前出现过的 A[i] + i
的最大值,则当前的数字就可以当作数对儿中的另一个数字,其减去当前坐标值再加上 mx 就可以更新结果 res 了,参见代码如下:
class Solution {
public:
int maxScoreSightseeingPair(vector<int>& A) {
int res = 0, n = A.size(), mx = 0;
for (int i = 0; i < n; ++i) {
res = max(res, mx + A[i] - i);
mx = max(mx, A[i] + i);
}
return res;
}
};
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