## 题目描述：

LeetCode 624. Maximum Distance in Arrays

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].

## Java代码：

public class Solution {
public int maxDistance(List<List<Integer>> arrays) {
int ans = 0;
int max = arrays.get(0).get(arrays.get(0).size() - 1);
int min = arrays.get(0).get(0);

for (int i = 1; i < arrays.size(); i++) {
ans = Math.max(ans, Math.abs(arrays.get(i).get(0) - max));
ans = Math.max(ans, Math.abs(arrays.get(i).get(arrays.get(i).size() - 1) - min));
max = Math.max(max, arrays.get(i).get(arrays.get(i).size() - 1));
min = Math.min(min, arrays.get(i).get(0));
}

return ans;
}
}

分别将max和min从maxMap，minMap中移除

利用maxMap.lastKey() - min，max - minMap.firstKey()更新答案

然后将max和min添加回maxMap与minMap

## Java代码：

public class Solution {
public int maxDistance(List<List<Integer>> arrays) {
TreeMap<Integer, Integer> maxMap = new TreeMap<>();
TreeMap<Integer, Integer> minMap = new TreeMap<>();
for (List<Integer> list : arrays) {
int min = list.get(0), max = list.get(list.size() - 1);
maxMap.put(max, maxMap.getOrDefault(max, 0) + 1);
minMap.put(min, minMap.getOrDefault(min, 0) + 1);
}
int ans = 0;
for (List<Integer> list : arrays) {
int min = list.get(0), max = list.get(list.size() - 1);
if (maxMap.put(max, maxMap.get(max) - 1) == 1) {
maxMap.remove(max);
}
if (minMap.put(min, minMap.get(min) - 1) == 1) {
minMap.remove(min);
}
ans = Math.max(ans, maxMap.lastKey() - min);
ans = Math.max(ans, max - minMap.firstKey());
maxMap.put(max, maxMap.getOrDefault(max, 0) + 1);
minMap.put(min, minMap.getOrDefault(min, 0) + 1);
}
return ans;
}
}

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