Weekly Contest 129--1021. Best Sightseeing Pair--Medium

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

1.思考

• 又是一开始就想到用暴力搜索，后来仔细一想就觉得不对；
• 然后开始研究A[i] + A[j] + i - j这个算是算式，就想到用A1、A2两个vector来分别存放A[i]+i、A[i]-i；
• 接着计算A1[i]+max(A2[i+1...len])，因为要一直用到max_element()这个函数，所以干脆就提前算好，A2[i]为从i到len-1的A2中最大的数；
• 最后直接用A1[i]+A2[i+1]，然后求出最大值。
  eg.
  index 0 1 2 3 4
  A 8 1 5 2 6
  A1 8 2 7 5 6
  (A2 8 0 3 -1 2)
  A2 8 3 3 2 2
  res 11 5 9 7 -
  
  

2.知识点

• *max_element()是直接返回最大值；
• *max_element()是返回最大值的位置，再减去A.begin()就得到index。
  
  

3.实现

class Solution {
public:
    int maxScoreSightseeingPair(vector<int>& A) {
        int len = A.size();
        vector<int>  A1, A2, score;
        for(int i=0; i<len; i++){
            A1.push_back(A[i]+i);
            A2.push_back(A[i]-i);
        }
        
        int d = *min_element(A2.begin(), A2.end());
        for(int i=len-1; i>=0; i--){
            d = A2[i]>d?A2[i]:d;
            A2[i] = d;
        }
        
        for(int i=0; i<len-1; i++){
            score.push_back(A1[i] + A2[i+1]);
        }
        
        int res = *max_element(score.begin(), score.end());
        return res;
    }
};