LeetCode-3：Longest Substring Without Repeating Characters (字符串最长子串)

题干：

Given a string, find the length of the longest substring without repeating characters.

给定一个字符串，求他的最长子串。

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: 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.

解析：

借助一个整数数组用于存储已出现的字符的下标，字符对应的整数值作为下标，数组的值为元素在字符串chars中的下标，如：asdf，表示为 charIndex[97] = 0.

1. 初始化一个整数数组charIndex，大小为128（ASCII码最多128个不同的字符），数组值初始化为-1。

2. 从左向右滑动检查字符是否在charIndex数组中出现过（charIndex数组的值不等于-1）。

3. 如果字符未出现过，计入长度，并判定当前字符串长度是否大于历史上最大长度。

4. 如果出现过，重新计算剩余长度curLen，并将出现位置之前到滑动窗口起始位置的值设置为-1

提交结果：

Runtime: 31 ms, faster than 98.64% of Java online submissions for Longest Substring Without Repeating Characters.

源码（Java）：

class Solution {
    public int lengthOfLongestSubstring(String s) {
        char[] chars = s.toCharArray();
        int maxLen = 0; //存储当前找到的最长字符串
        int curLen = 0; //存储当前不重复字符串长度
        int[] charIndex = new int[128]; //ASCII一共128个字符
        //初始化成-1
        for (int i=0;i<charIndex.length;i++){
            charIndex[i] = -1;
        }
        int pre = 0;
        for (int i=0;i<chars.length;i++){
            int index = chars[i];
            if (charIndex[index] >= 0){ //已经存在这个字符了
                //判断如果长度是大于maxLen，则替换，否则不操作
                if (curLen>maxLen){ // maxLen = curLen>maxLen?curLen:maxLen;
                    maxLen = curLen;
                }
                //重新计算长度
                curLen = i - charIndex[index];
                //将之前的值置成0
                int tmp = pre;
                pre = charIndex[index] + 1; //记录pre值
                for (int j=charIndex[index];j>=tmp;j--){
                    charIndex[(int)chars[j]] = -1;
                }
                charIndex[index] = i; //记录字符串出现过
            } else {
                //不存在已经出现的字符串
                curLen ++;
                if (curLen>maxLen){
                    maxLen = curLen;
                }
                charIndex[index] = i;
            }
        }
        return maxLen;
    }
}