LeetCode練習-字符串-longest-substring-without-repeating-characters

題目描述
Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

思路：滑動窗口

eg: abcabcbb，

a

ab

abc

bca

cab

......

類似與deque，若子串中有當前字符，則把子串中當前字符及前邊的字符都刪除並清空對應的標誌位，最後把當前字符添加到尾部；

若子串中沒有當前字符，則直接把當前字符添加到子串尾部，並設置對應的標誌位。

代碼：

/*longest-substring-without-repeating-characters*/

int lengthOfLongestSubstring(string s);
bool charIsInSubString(deque<char> d, char c);//判斷當前字符是否已在子串中
int lengthOfLongestSubstring_2(string s);//改進了判斷當前字符是否已在子串中。
int lengthOfLongestSubstring_3(string s);//改進了deque，用兩個變量begin和end來模擬deque。

int main(){
	//char c = 'a';
	//deque<char> d;
	//d.push_back('a');
	//d.push_back('b');
	//cout << charIsInSubString(d, c) << endl;

	string s = "wlrbbmqbhcdarzowkkyhiddqscdxrjmowfrxsjybldbefsarcbynecdyggxxpklorellnmpapqfwkhopkmco";
	//string s= "abcabcbb";
	//string s= "bbbbb";
	cout << lengthOfLongestSubstring_3(s) << endl;
	return 0;
}

int lengthOfLongestSubstring_3(string s){
	bool flag[255] = {0};
	int max=0;
	int count = 0;
	int i;
	int begin=0, end=0;
	for(i = 0; i < s.length(); ++i){
		if(!flag[s[i]]){
			flag[s[i]] = true;
			++end;
			++count;
			if(count > max)
				max = count;
		}
		else{
			int j;
			for(j = begin; s[j] != s[i]; ++j){
				flag[s[j]] = false;
				++begin;
				--count;
			}
			++begin;
			--count;
			
			++end;
			++count;
			if(count > max)
				max = count;
		}
	}
	return max;
}

int lengthOfLongestSubstring_2(string s){//運行時間8ms，內存492k
	bool flag[255] = {0};
	
	//ostream_iterator<char> out_it(cout, " ");
	int max=0;
	deque<char> d;
	int count = 0;
	int i;
	for(i = 0; i < s.length(); ++i){
		if(!flag[s[i]]){
			flag[s[i]] = true;
			d.push_back(s[i]);
			++count;
			if(count > max)
				max = count;
		}
		else{
			deque<char>::iterator dIter = d.begin();
			for(; *dIter != s[i]; ++dIter){
				flag[d.front()] = false;
				d.pop_front();
				--count;
			}
			d.pop_front();
			--count;
			
			d.push_back(s[i]);
			++count;
			if(count > max)
				max = count;
		}
		//copy(d.begin(), d.end(), out_it);
		//cout << " " << "count: " << count << " max: "<< max << endl;
	}
	return max;
}
int lengthOfLongestSubstring(string s){//運行時間20ms，內存492k
	ostream_iterator<char> out_it(cout, " ");
	int max=0;
	deque<char> d;
	int count = 0;
	int i;
	for(i = 0; i < s.length(); ++i){
		if(!charIsInSubString(d, s[i])){
			d.push_back(s[i]);
			++count;
			if(count > max)
				max = count;
		}
		else{
			deque<char>::iterator dIter = d.begin();
			for(; *dIter != s[i]; ++dIter){
				d.pop_front();
				--count;
			}
			d.pop_front();
			--count;
			d.push_back(s[i]);
			++count;
			if(count > max)
				max = count;
		}
		//copy(d.begin(), d.end(), out_it);
		//cout << " " << "count: " << count << " max: "<< max << endl;
	}
	return max;
}

bool charIsInSubString(deque<char> d, char c){
	deque<char>::iterator dIter;
	dIter = find(d.begin(), d.end(), c);
	if(dIter == d.end())
		return false;
	else
		return true;
}