Given m arrays, and each array is sorted in ascending order. Now you can pick up two integers from two different arrays (each array picks one) and calculate the distance. We define the distance between two integers a and b to be their absolute difference |a-b|. Your task is to find the maximum distance.
给定m个数组，均为升序。可于两个数组中各取一个数来计算其距离，这里距离被定义为两数差的绝对值。算法要求返回最大的距离。
Example 1:
Input:
[[1,2,3],
[4,5],
[1,2,3]]
Output: 4
Explanation:
One way to reach the maximum distance 4 is to pick 1 in the first or third array and pick 5 in the second array.
Note:
- Each given array will have at least 1 number. There will be at least two non-empty arrays.
- The total number of the integers in all the m arrays will be in the range of [2, 10000].
- The integers in the m arrays will be in the range of [-10000, 10000].
思路
升序排列，开头最小末尾最大，只需扫描数组记录最小左值和最大右值之差，并确保不在同一行即可。
class Solution {
public:
int maxDistance(vector<vector<int>>& arrays) {
int left = arrays[0][0], right = arrays[0].back(), ans = 0;
for(int i = 1; i < arrays.size(); i++){
ans = max(ans, max(abs(arrays[i][0] - right), abs(arrays[i].back() - left)));
left = min(left, arrays[i][0]);
right = max(right, arrays[i].back());
}
return ans;
}
};
