目录链接：

力扣编程题-解法汇总_分享+记录-CSDN博客

GitHub同步刷题项目：

https://github.com/September26/java-algorithms

原题链接：力扣

描述：

给定一个放有字母和数字的数组，找到最长的子数组，且包含的字母和数字的个数相同。

返回该子数组，若存在多个最长子数组，返回左端点下标值最小的子数组。若不存在这样的数组，返回一个空数组。

示例 1:

输入: ["A","1","B","C","D","2","3","4","E","5","F","G","6","7","H","I","J","K","L","M"]输出: ["A","1","B","C","D","2","3","4","E","5","F","G","6","7"]

示例 2:

输入: ["A","A"]输出: []

提示：

• array.length <= 100000

解题思路：

* 解题思路：
* 记录每个位置charNum-intNum的值，按照这个值来分组，满足统一条件的两个位置相减，就是符合条件的子数组。
* 构建一个Map>的结构，某个位置charNum-intNum的结果，value为满足这个结果的所有位置的集合。
* 这里要特别注意一下diff=0的情况，0的话时候额外填充一个0。
* 然后遍历map的value，取出集合，集合的最后一位减去第一位，就是满足条件的长度。求最长。

代码：

public class MianShi1705 {public String[] findLongestSubarray(String[] array) {Map> map = new HashMap<>();int intNum = 0;int charNum = 0;for (int i = 0; i < array.length; i++) {String s = array[i];char c = s.toCharArray()[0];if ((c >= 'A' && c <= 'Z') || c >= 'a' && c <= 'z') {charNum++;} else {intNum++;}int diff = charNum - intNum;List integers = obtainQueue(map, diff);if (diff == 0 && integers.size() == 0) {integers.add(0);}integers.add(i + 1);}int maxLength = 0;int maxStart = 0;int maxEnd = 0;for (int key : map.keySet()) {List list = map.get(key);if (list.size() == 0) {continue;}int length;if (list.size() == 1) {continue;}length = list.get(list.size() - 1) - list.get(0);if (length > maxLength || (length == maxLength && list.get(0) < maxStart)) {maxLength = length;maxStart = list.get(0);maxEnd = list.get(list.size() - 1);}}if (maxLength == 0) {return new String[0];}String[] strings = Arrays.copyOfRange(array, maxStart, maxEnd);return strings;}private List obtainQueue(Map> map, int key) {List integers = map.get(key);if (integers == null) {integers = new ArrayList<>();map.put(key, integers);}return integers;}}