https://leetcode-cn.com/problems/find-longest-subarray-lcci/
给定一个放有字符和数字的数组,找到最长的子数组,且包含的字符和数字的个数相同。
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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/find-longest-subarray-lcci
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将数字和字母分别转换成1和-1
计算前缀和
如果前缀和==0
那么就计算区间长度
如果前缀和出现相同的数字
计算区间长度
返回最长的区间长度的内容
执行用时: 200 ms , 在所有 Python3 提交中击败了 49.01% 的用户 内存消耗: 31.3 MB , 在所有 Python3 提交中击败了 87.42% 的用户
class Solution:
def findLongestSubarray(self, array: List[str]) -> List[str]:
length = array.__len__()
prefix_sum = [0] * length # 记录前缀和
left, right = 0, 0 # 记录返回的数组下标
sub_dict = {} # 记录出现过的非0前缀和 首次出现的下标
# 字母标记为1,数字标记为-1
for i in range(length):
prefix_sum[i] = 1 if 'A' <= array[i] <= 'z' else -1
# 记录前缀和
for i in range(1, length):
prefix_sum[i] += prefix_sum[i-1]
for index, _sum in enumerate(prefix_sum):
# 前缀和为0的,说明下标0 到 index的字母数字出现次数相等,下标包含0和index
if _sum == 0:
if right - left < index:
left = 0
right = index + 1
else:
# 前缀和出现过,计算等钱index和首次出现的下标差值,用差值和当前记录的最长字符串长度再做比较,如果大于
# 当前记录的范围长度,则进行记录
if _sum in sub_dict:
if right - left < index - sub_dict[_sum]:
left = sub_dict[_sum] + 1
right = index + 1
else:
# 前缀和未出现过,则记录首次出现的下标
sub_dict[_sum] = index
return array[left: right]