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.

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].

給m個數組， 每個數組按升序排列。現在你可以從2個不同的數組中各取1個數字，計算他們的距離，距離是兩個數值差的絕對值。任務是找出最大距離。

註意要從不同數組中取數，那麽即使某個數組的首尾差距很大，也不行。

解法1：最大堆和最小堆。空間復雜度高。

解法2：用min_val和max_val表示當前遍歷到的數組中最小的首元素和最大的尾元素，當遍歷到一個新的數組時，計算新數組尾元素和min_val絕對差以及max_val和新數組首元素的絕對差，取較大值和result比較來更新結果，最後返回result。

Java:

public class Solution {
    public int maxDistance(int[][] list) {
        int res = 0, min_val = list[0][0], max_val = list[0][list[0].length - 1];
        for (int i = 1; i < list.length; i++) {
            res = Math.max(res, Math.max(Math.abs(list[i][list[i].length - 1] - min_val), Math.abs(max_val - list[i][0])));
            min_val = Math.min(min_val, list[i][0]);
            max_val = Math.max(max_val, list[i][list[i].length - 1]);
        }
        return res;
    }
}　　

Python:

# Time:  O(n)
# Space: O(1)
class Solution(object):
    def maxDistance(self, arrays):
        """
        :type arrays: List[List[int]]
        :rtype: int
        """
        result, min_val, max_val = 0,  arrays[0][0], arrays[0][-1]
        for i in xrange(1, len(arrays)):
            result = max(result,  max(max_val - arrays[i][0], arrays[i][-1] - min_val))
            min_val = min(min_val, arrays[i][0])
            max_val = max(max_val, arrays[i][-1])
        return result　　

C++:

class Solution {
public:
    int maxDistance(vector<vector<int>>& arrays) {
        priority_queue<pair<int, int>> mx, mn;
        for (int i = 0; i < arrays.size(); ++i) {
            mn.push({-arrays[i][0], i});
            mx.push({arrays[i].back(), i});
        }
        auto a1 = mx.top(); mx.pop();
        auto b1 = mn.top(); mn.pop();
        if (a1.second != b1.second) return a1.first + b1.first;
        return max(a1.first + mn.top().first, mx.top().first + b1.first);
    }
};

C++:

class Solution {
public:
    int maxDistance(vector<vector<int>>& arrays) {
        int res = 0, start = arrays[0][0], end = arrays[0].back();
        for (int i = 1; i < arrays.size(); ++i) {
            res = max(res, max(abs(arrays[i].back() - start), abs(end - arrays[i][0])));
            start = min(start, arrays[i][0]);
            end = max(end, arrays[i].back());
        }
        return res;
    }
};

[LeetCode] 624. Maximum Distance in Arrays 數組中的最大距離