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每日算法系列【LeetCode 面试题 17.05】字母与数字

水沐由之 2022-07-27 阅读 24


题目描述

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

题解

每日算法系列【LeetCode 面试题 17.05】字母与数字_c++

代码

c++


class Solution {
public:
vector<string> findLongestSubarray(vector<string>& array) {
int n = array.size();
unordered_map<int, int> mp;
mp[0] = -1;
int cnt = 0, l = 0, r = 0;
for (int i = 0; i < n; ++i) {
cnt += isdigit(array[i][0]) ? 1 : -1;
if (mp.find(cnt) != mp.end()) {
if (r-l < i-mp[cnt]) {
l = mp[cnt] + 1;
r = i + 1;
}
} else {
mp[cnt] = i;
}
}
return vector<string>(array.begin()+l, array.begin()+r);
}
};


python


class Solution:
def findLongestSubarray(self, array: List[str]) -> List[str]:
mp = {0: -1}
cnt, l, r = 0, 0, 0
for i, s in enumerate(array):
cnt += 1 if s[:1].isdigit() else -1
if cnt in mp:
if r-l < i-mp[cnt]:
l = mp[cnt] + 1
r = i + 1
else:
mp[cnt] = i
return array[l:r]


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