Longest Substring Without Repeating Characters(最长不重复子序列求解)

问题描述：

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.

给出一个字符串，找到它的最长无重复子序列长度。

思路1

从下标0开始，依次查找以当前下标开始的最长不重复子序列。

举例说明：

C++代码实现：

int GetMaxLength(string& sTemp, int startIndex)
{map<char, int> mapChar;int max = 0;for (int i = startIndex; i < sTemp.length(); ++i){if (mapChar.find(sTemp[i]) != mapChar.end())return max;mapChar[sTemp[i]] = i;++max;}return max;
}int lengthOfLongestSubstring(string s)
{int max = 0;for (int i = 0; i < s.length(); ++i){int temp = GetMaxLength(s, i);if (temp > max)max = temp;}return max;
}

思路2

设当前检测的子序列，下标最小值为m，最大值为n（即子串strInput[m,n]），则当前检测序列的长度为n-m+1。

如果该序列中没有重复元素，那么它的长度可以作为最长无重复子序列长度的一个候选值（如果后来有的序列长度超过了它，就替换掉）

我们设当前检测序列的起始下标为now，遍历整个字符串，遍历中当前下标为i；如果可以保证从now到i中没有重复元素，遍历一次，就可以利用i-now+1找到最长无重复子序列。

使now从0开始，后移遍历下标i，直到strInput[now,i]中出现重复的字符，更新now的值，将now移动到重复字符第一次出现的下标后即可。

举例说明：

可以将字符和下标存入哈希表中，这样在查找新加入的字符是否在原始检索串中存在时，时间复杂度为1。

C++代码实现：

int lengthOfLongestSubstring(string s)
{int maxLength = 0;int now = 0;map<char, int> charMap;for (int i = 0; i < s.length(); ++i){if (charMap.find(s[i]) != charMap.end()){now = charMap[s[i]] + 1 > now ? charMap[s[i]] + 1 : now;}if (i - now + 1>maxLength){maxLength = i - now + 1;}charMap[s[i]] = i;}return maxLength;
}

思路2优化

由于在C++中字符和数字转换非常简单，所以，可以用一个数组来替代哈希表：

C++代码实现：

int lengthOfLongestSubstring(string s)
{int maxLength = 0;int now = 0;int charMap[256];for (int i = 0; i < 256; ++i){charMap[i] = -1;}const char *p = s.c_str();for (int i = 0; i < s.length(); ++i){if (charMap[p[i]] != -1){now = charMap[p[i]] + 1 > now ? charMap[p[i]] + 1 : now;}if (i - now + 1>maxLength){maxLength = i - now + 1;}charMap[p[i]] = i;}return maxLength;
}

转载于:https://www.cnblogs.com/zhaogl/p/6364654.html

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