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.
給定m個(gè)數(shù)組，均為升序?？捎趦蓚€(gè)數(shù)組中各取一個(gè)數(shù)來(lái)計(jì)算其距離，這里距離被定義為兩數(shù)差的絕對(duì)值。算法要求返回最大的距離。

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:

1. Each given array will have at least 1 number. There will be at least two non-empty arrays.
2. The total number of the integers in all the m arrays will be in the range of [2, 10000].
3. The integers in the m arrays will be in the range of [-10000, 10000].

思路
升序排列，開頭最小末尾最大，只需掃描數(shù)組記錄最小左值和最大右值之差，并確保不在同一行即可。

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;
    }
};