leetcode 1371. 每个元音包含偶数次的最长子字符串
leetcode 1371. 每个元音包含偶数次的最长子字符串
题目详情
题目链接
给你一个字符串 s ，请你返回满足以下条件的最长子字符串的长度：每个元音字母，即 ‘a’，‘e’，‘i’，‘o’，‘u’ ，在子字符串中都恰好出现了偶数次。
- 示例 1：
输入：s = “eleetminicoworoep”
输出：13
解释：最长子字符串是 “leetminicowor” ，它包含 e，i，o 各 2 个，以及 0 个 a，u 。- 示例 2：
输入：s = “leetcodeisgreat”
输出：5
解释：最长子字符串是 “leetc” ，其中包含 2 个 e 。- 示例 3：
输入：s = “bcbcbc”
输出：6
解释：这个示例中，字符串 “bcbcbc” 本身就是最长的，因为所有的元音 a，e，i，o，u 都出现了 0 次。提示：
- 1 <= s.length <= 5 x 10^5
- s 只包含小写英文字母。
我的代码
class Solution {
public:
int longestSubstring(const string &s, int start, int end, map<char, int> &count) {
if (count['a'] % 2) {
auto temp = count;
bool isFind = false;
int len1 = 0, len2 = 0;
for (int i = start; i <= end; ++i) {
switch (s[i]) {
case 'a': --count['a'], isFind = true; break;
case 'e': --count['e']; break;
case 'i': --count['i']; break;
case 'o': --count['o']; break;
case 'u': --count['u']; break;
default: break;
}
if (isFind) {
len1 = longestSubstring(s, i + 1, end, count);
break;
}
}
isFind = false;
count = temp;
for (int i = end; i >= start; --i) {
switch (s[i]) {
case 'a': --count['a'], isFind = true; break;
case 'e': --count['e']; break;
case 'i': --count['i']; break;
case 'o': --count['o']; break;
case 'u': --count['u']; break;
default: break;
}
if (isFind) {
len2 = longestSubstring(s, start, i - 1, count);
break;
}
}
return max(len1, len2);
} else if (count['e'] % 2) {
auto temp = count;
bool isFind = false;
int len1 = 0, len2 = 0;
for (int i = start; i <= end; ++i) {
switch (s[i]) {
case 'a': --count['a']; break;
case 'e': --count['e'], isFind = true; break;
case 'i': --count['i']; break;
case 'o': --count['o']; break;
case 'u': --count['u']; break;
default: break;
}
if (isFind) {
len1 = longestSubstring(s, i + 1, end, count);
break;
}
}
isFind = false;
count = temp;
for (int i = end; i >= start; --i) {
switch (s[i]) {
case 'a': --count['a']; break;
case 'e': --count['e'], isFind = true; break;
case 'i': --count['i']; break;
case 'o': --count['o']; break;
case 'u': --count['u']; break;
default: break;
}
if (isFind) {
len2 = longestSubstring(s, start, i - 1, count);
break;
}
}
return max(len1, len2);
} else if (count['i'] % 2) {
auto temp = count;
bool isFind = false;
int len1 = 0, len2 = 0;
for (int i = start; i <= end; ++i) {
switch (s[i]) {
case 'a': --count['a']; break;
case 'e': --count['e']; break;
case 'i': --count['i'], isFind = true; break;
case 'o': --count['o']; break;
case 'u': --count['u']; break;
default: break;
}
if (isFind) {
len1 = longestSubstring(s, i + 1, end, count);
break;
}
}
isFind = false;
count = temp;
for (int i = end; i >= start; --i) {
switch (s[i]) {
case 'a': --count['a']; break;
case 'e': --count['e']; break;
case 'i': --count['i'], isFind = true; break;
case 'o': --count['o']; break;
case 'u': --count['u']; break;
default: break;
}
if (isFind) {
len2 = longestSubstring(s, start, i - 1, count);
break;
}
}
return max(len1, len2);
} else if (count['o'] % 2) {
auto temp = count;
bool isFind = false;
int len1 = 0, len2 = 0;
for (int i = start; i <= end; ++i) {
switch (s[i]) {
case 'a': --count['a']; break;
case 'e': --count['e']; break;
case 'i': --count['i']; break;
case 'o': --count['o'], isFind = true; break;
case 'u': --count['u']; break;
default: break;
}
if (isFind) {
len1 = longestSubstring(s, i + 1, end, count);
break;
}
}
isFind = false;
count = temp;
for (int i = end; i >= start; --i) {
switch (s[i]) {
case 'a': --count['a']; break;
case 'e': --count['e']; break;
case 'i': --count['i']; break;
case 'o': --count['o'], isFind = true; break;
case 'u': --count['u']; break;
default: break;
}
if (isFind) {
len2 = longestSubstring(s, start, i - 1, count);
break;
}
}
return max(len1, len2);
} else if (count['u'] % 2) {
auto temp = count;
bool isFind = false;
int len1 = 0, len2 = 0;
for (int i = start; i <= end; ++i) {
switch (s[i]) {
case 'a': --count['a']; break;
case 'e': --count['e']; break;
case 'i': --count['i']; break;
case 'o': --count['o']; break;
case 'u': --count['u'], isFind = true; break;
default: break;
}
if (isFind) {
len1 = longestSubstring(s, i + 1, end, count);
break;
}
}
isFind = false;
count = temp;
for (int i = end; i >= start; --i) {
switch (s[i]) {
case 'a': --count['a']; break;
case 'e': --count['e']; break;
case 'i': --count['i']; break;
case 'o': --count['o']; break;
case 'u': --count['u'], isFind = true; break;
default: break;
}
if (isFind) {
len2 = longestSubstring(s, start, i - 1, count);
break;
}
}
return max(len1, len2);
} else {
return end + 1 - start;
}
}
int findTheLongestSubstring(string s) {
int len = s.size();
map<char, int> count;
count['a'] = 0, count['e'] = 0, count['i'] = 0, count['o'] = 0, count['u'] = 0;
for (const auto &c: s) {
switch (c) {
case 'a': ++count['a']; break;
case 'e': ++count['e']; break;
case 'i': ++count['i']; break;
case 'o': ++count['o']; break;
case 'u': ++count['u']; break;
default: break;
}
}
return longestSubstring(s, 0, len - 1, count);
}
};
我的成绩
执行结果：超出时间限制。
一些想法
本道题我使用了递归，单纯的递归果然是无法解决中等难度的题的。。。
执行用时为 32 ms 的范例
#include 思考
不会做。。。有时间把范例看一下