题解 | #最长无重复子数组#

解题思路：
1. 边界条件1，数组长度为1，输出1
边界条件2，数组不重复，输出数组长度
2. 双指针，起始指针start_i, 和结尾指针end_i,
arr[start_i: end_i] 保证不重复
且 arr[start_i: end_i+1] 重复
3. 更新的双指针，找出最大长度

class Solution:
    def _find_end_idx(self, arr, start_i, end_i, curr_i):
        """找到终点"""
        end_idx = -1
        tmp_set = set(arr[start_i: curr_i+1])
        for i in range(curr_i+1, end_i+1):
            if arr[i] in tmp_set:
                end_idx = i
                break
            tmp_set.add(arr[i])
        if end_idx == -1:
            return end_i
        return end_idx

    def _find_start_idx(self, arr, start_i, end_i, curr_num):
        """找到起始点"""
        for i in range(start_i, end_i):
            if curr_num == arr[i]:
                return i + 1

    def maxLength(self , arr ):
        # write code here
        # 边界,长度为1
        if len(arr) == 1:
            return 1

        # 边界，均无重复
        if len(arr) == len(set(arr)):
            return len(arr)

        # 双指针
        # 1. 找到第一个非重复的数组终点
        end_idx, start_idx, max_len = -1, 0, -1
        tmp_set = {arr[0]}
        for i in range(1, len(arr)):
            if arr[i] in tmp_set:
                end_idx = i
                max_len = i
                break
            tmp_set.add(arr[i])

        # 2. 开始循环执行，找到起点，更新长度, 找终点
        while True:

            if end_idx == len(arr) - 1:
                break

            start_idx = self._find_start_idx(arr, start_idx, end_idx, arr[end_idx])
            if start_idx > len(arr) - 1:
                break
            max_len = max(max_len, end_idx - start_idx + 1)

            end_idx = self._find_end_idx(arr, start_idx, len(arr) - 1, end_idx)
            max_len = max(max_len, end_idx - start_idx)

        return max_len