Given a string, find the length of the longest substring without repeating characters.
Examples:
Given "abcabcbb", the answer is "abc",
which the length is 3.
Given "bbbbb", the answer is "b",
with the length of 1.
Given "pwwkew", the answer is "wke",
with the length of 3. Note that the answer must be a substring, "pwke" is
a subsequence and not a substring.
時間複雜度O(n^2)
#include <iostream>
#include <string>
using namespace std;
class Solution {
public:
int lengthOfLongestSubstring(string s) {
size_t found;
int max_len = s.empty() ? 0 : 1;
size_t prev = 0, next = 1;
int len = 1;
while (next < s.size()) {
found = find(s.at(next), s, prev, next);
if (found == string::npos) {
++len;
++next;
} else {
if (len > max_len) {
max_len = len;
}
prev = found + 1;
len = next - prev + 1;
++next;
}
}
// 防止最後一個窗口沒有將較大的len賦值給max_len,同時考慮s爲空的情形
if (len > max_len && !s.empty()) {
max_len = len;
}
return max_len;
}
private:
size_t find(char ch, string s, size_t start_pos, size_t end_pos) {
size_t pos = start_pos;
while (pos < end_pos) {
if (s.at(pos) == ch) {
return pos;
}
pos++;
}
return string::npos;
}
};
int main() {
string str1 = "abcabcbb";
string str2 = "bbbbb";
string str3 = "pwwkew";
string str4 = "au";
string str5 = "a";
string str6;
Solution s;
cout << s.lengthOfLongestSubstring(str1) << endl
<< s.lengthOfLongestSubstring(str2) << endl
<< s.lengthOfLongestSubstring(str3) << endl
<< s.lengthOfLongestSubstring(str4) << endl
<< s.lengthOfLongestSubstring(str5) << endl
<< s.lengthOfLongestSubstring(str6) << endl;
return 0;
}