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feat: add solutions to lc problem: No.0624
No.0624.Maximum Distance in Arrays
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‎solution/0600-0699/0624.Maximum Distance in Arrays/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：维护最大值和最小值**
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我们注意到，最大距离一定是两个数组中的一个最大值和另一个最小值之间的距离。因此，我们可以维护两个变量，分别表示当前数组中的最大值和最小值，然后遍历数组，更新最大距离，同时更新最大值和最小值。
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遍历结束后，即可得到最大距离。
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时间复杂度 $O(m)$，空间复杂度 $O(1)$。其中 $m$ 为数组的个数。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maxDistance(self, arrays: List[List[int]]) -> int:
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ans = 0
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mi, mx = arrays[0][0], arrays[0][-1]
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for arr in arrays[1:]:
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a, b = abs(arr[0] - mx), abs(arr[-1] - mi)
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ans = max(ans, a, b)
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mi = min(mi, arr[0])
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mx = max(mx, arr[-1])
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int maxDistance(List<List<Integer>> arrays) {
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int ans = 0;
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int mi = arrays.get(0).get(0);
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int mx = arrays.get(0).get(arrays.get(0).size() - 1);
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for (int i = 1; i < arrays.size(); ++i) {
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var arr = arrays.get(i);
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int a = Math.abs(arr.get(0) - mx);
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int b = Math.abs(arr.get(arr.size() - 1) - mi);
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ans = Math.max(ans, Math.max(a, b));
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mi = Math.min(mi, arr.get(0));
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mx = Math.max(mx, arr.get(arr.size() - 1));
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxDistance(vector<vector<int>>& arrays) {
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int ans = 0;
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int mi = arrays[0][0], mx = arrays[0][arrays[0].size() - 1];
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for (int i = 1; i < arrays.size(); ++i) {
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auto& arr = arrays[i];
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int a = abs(arr[0] - mx), b = abs(arr[arr.size() - 1] - mi);
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ans = max({ans, a, b});
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mi = min(mi, arr[0]);
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mx = max(mx, arr[arr.size() - 1]);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxDistance(arrays [][]int) (ans int) {
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mi, mx := arrays[0][0], arrays[0][len(arrays[0])-1]
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for _, arr := range arrays[1:] {
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a, b := abs(arr[0]-mx), abs(arr[len(arr)-1]-mi)
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ans = max(ans, max(a, b))
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mi = min(mi, arr[0])
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mx = max(mx, arr[len(arr)-1])
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0600-0699/0624.Maximum Distance in Arrays/README_EN.md

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### **Python3**
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```python
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class Solution:
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def maxDistance(self, arrays: List[List[int]]) -> int:
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ans = 0
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mi, mx = arrays[0][0], arrays[0][-1]
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for arr in arrays[1:]:
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a, b = abs(arr[0] - mx), abs(arr[-1] - mi)
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ans = max(ans, a, b)
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mi = min(mi, arr[0])
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mx = max(mx, arr[-1])
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int maxDistance(List<List<Integer>> arrays) {
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int ans = 0;
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int mi = arrays.get(0).get(0);
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int mx = arrays.get(0).get(arrays.get(0).size() - 1);
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for (int i = 1; i < arrays.size(); ++i) {
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var arr = arrays.get(i);
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int a = Math.abs(arr.get(0) - mx);
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int b = Math.abs(arr.get(arr.size() - 1) - mi);
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ans = Math.max(ans, Math.max(a, b));
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mi = Math.min(mi, arr.get(0));
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mx = Math.max(mx, arr.get(arr.size() - 1));
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxDistance(vector<vector<int>>& arrays) {
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int ans = 0;
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int mi = arrays[0][0], mx = arrays[0][arrays[0].size() - 1];
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for (int i = 1; i < arrays.size(); ++i) {
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auto& arr = arrays[i];
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int a = abs(arr[0] - mx), b = abs(arr[arr.size() - 1] - mi);
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ans = max({ans, a, b});
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mi = min(mi, arr[0]);
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mx = max(mx, arr[arr.size() - 1]);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maxDistance(arrays [][]int) (ans int) {
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mi, mx := arrays[0][0], arrays[0][len(arrays[0])-1]
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for _, arr := range arrays[1:] {
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a, b := abs(arr[0]-mx), abs(arr[len(arr)-1]-mi)
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ans = max(ans, max(a, b))
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mi = min(mi, arr[0])
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mx = max(mx, arr[len(arr)-1])
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0600-0699/0624.Maximum Distance in Arrays/Solution.cpp

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class Solution {
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public:
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int maxDistance(vector<vector<int>>& arrays) {
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int ans = 0;
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int mi = arrays[0][0], mx = arrays[0][arrays[0].size() - 1];
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for (int i = 1; i < arrays.size(); ++i) {
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auto& arr = arrays[i];
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int a = abs(arr[0] - mx), b = abs(arr[arr.size() - 1] - mi);
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ans = max({ans, a, b});
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mi = min(mi, arr[0]);
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mx = max(mx, arr[arr.size() - 1]);
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}
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return ans;
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}
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};

‎solution/0600-0699/0624.Maximum Distance in Arrays/Solution.go

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func maxDistance(arrays [][]int) (ans int) {
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mi, mx := arrays[0][0], arrays[0][len(arrays[0])-1]
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for _, arr := range arrays[1:] {
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a, b := abs(arr[0]-mx), abs(arr[len(arr)-1]-mi)
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ans = max(ans, max(a, b))
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mi = min(mi, arr[0])
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mx = max(mx, arr[len(arr)-1])
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}
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return ans
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}

‎solution/0600-0699/0624.Maximum Distance in Arrays/Solution.java

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class Solution {
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public int maxDistance(List<List<Integer>> arrays) {
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int ans = 0;
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int mi = arrays.get(0).get(0);
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int mx = arrays.get(0).get(arrays.get(0).size() - 1);
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for (int i = 1; i < arrays.size(); ++i) {
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var arr = arrays.get(i);
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int a = Math.abs(arr.get(0) - mx);
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int b = Math.abs(arr.get(arr.size() - 1) - mi);
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ans = Math.max(ans, Math.max(a, b));
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mi = Math.min(mi, arr.get(0));
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mx = Math.max(mx, arr.get(arr.size() - 1));
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}
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return ans;
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}
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}

‎solution/0600-0699/0624.Maximum Distance in Arrays/Solution.py

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class Solution:
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def maxDistance(self, arrays: List[List[int]]) -> int:
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ans = 0
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mi, mx = arrays[0][0], arrays[0][-1]
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for arr in arrays[1:]:
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a, b = abs(arr[0] - mx), abs(arr[-1] - mi)
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ans = max(ans, a, b)
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mi = min(mi, arr[0])
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mx = max(mx, arr[-1])
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return ans

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