最简洁的代码 求字符串中无重复的子串

这道 题是leetcode 上面的 一道题。常规解法需要两层循环。这种解法不仅代码简洁，而且只需要一层循环。降低了复杂度。一起欣赏一下作者的代码。

原题：

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.
*/

实现代码：简直简洁有木有！！！

/**
 * 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;
}