這道 題是leetcode 上面的 一道題。常規解法需要兩層循環。這種解法不僅代碼簡潔，而且只需要一層循環。降低了複雜度。一起欣賞一下作者的代碼。
原題：
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.
解法：
/**
* Solution (DP, O(n)):
*
* Assume L[i] = s[m...i], denotes the longest substring without repeating
* characters that ends up at s[i], and we keep a hashmap for every
* characters between m ... i, while storing <character, index> in the
* hashmap.
* We know that each character will appear only once.
* Then to find s[i+1]:
* 1) if s[i+1] does not appear in hashmap
* we can just add s[i+1] to hash map. and L[i+1] = s[m...i+1]
* 2) if s[i+1] exists in hashmap, and the hashmap value (the index) is k
* let m = max(m, k), then L[i+1] = s[m...i+1], we also need to update
* entry in hashmap to mark the latest occurency of s[i+1].
*
* Since we scan the string for only once, and the 'm' will also move from
* beginning to end for at most once. Overall complexity is O(n).
*
* If characters are all in ASCII, we could use array to mimic hashmap.
*/
實現代碼：簡直簡潔有木有！！！
/**
* Solution (DP, O(n)):
*
* Assume L[i] = s[m...i], denotes the longest substring without repeating
* characters that ends up at s[i], and we keep a hashmap for every
* characters between m ... i, while storing <character, index> in the
* hashmap.
* We know that each character will appear only once.
* Then to find s[i+1]:
* 1) if s[i+1] does not appear in hashmap
* we can just add s[i+1] to hash map. and L[i+1] = s[m...i+1]
* 2) if s[i+1] exists in hashmap, and the hashmap value (the index) is k
* let m = max(m, k), then L[i+1] = s[m...i+1], we also need to update
* entry in hashmap to mark the latest occurency of s[i+1].
*
* Since we scan the string for only once, and the 'm' will also move from
* beginning to end for at most once. Overall complexity is O(n).
*
* If characters are all in ASCII, we could use array to mimic hashmap.
*/
int lengthOfLongestSubstring(string s) {
// for ASCII char sequence, use this as a hashmap
vector<int> charIndex(256, -1);
int longest = 0, m = 0;
for (int i = 0; i < s.length(); i++) {
m = max(charIndex[s[i]] + 1, m); // automatically takes care of -1 case
charIndex[s[i]] = i;
longest = max(longest, i - m + 1);
}
return longest;
}