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:
题目的意思是:给你一个数组,找出最大的sightseeing pair对,即求下面的最大值:
A[i] + A[j] + i - j
这道题我暴力破解了一下,果然超时了,后面参考了一下别人的实现,global_max存放的就是遍历过程中的全局最大值,也是最后的结果,max_val存放的是第一个值,ma_inx就是对应的索引了。这里如果要更新max_val的话,要满足当前的值A[i]>=max_val+max_inx-i就行了,如果没有想到这个点就做不出来了。
class Solution:
def maxScoreSightseeingPair(self, A: List[int]) -> int:
n=len(A)
max_val=A[0]
max_inx=0
global_max=-1
for i in range(1,n):
if(A[i]+max_val+max_inx-i>global_max):
global_max=A[i]+max_val+max_inx-i
if(max_val+max_inx-i<=A[i]):
max_val=A[i]
max_inx=i
return global_max
[LeetCode] Easy Python solution beats 98%