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

Example 1:

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

Example 2:

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

Example 3:

Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

Constraints:

• 0 <= s.length <= 5 * 104
• s consists of English letters, digits, symbols and spaces.

給定一個字串 s

要求寫一個演算法找出子字串中不包含重複字元的最長長度

最直接的想法是

當還沒遇到重複字元時就累加長度。遇到重複字元時，就從上一次開始累加的位置start 右移一位， 也就是start+1 當作下一個開始做累計

具體作法如下：

初始化最大長度 maxLen = 0, start = 0, visitMap

然後逐步檢查每個字元 s[i]

每次先檢查 s[i] 是否已經存在 visitMap

如果s[i] 存在 visitMap 且 visitMap[s[i]] ≥ start 更新 start = vistMap[s[i]] +1

如果s[i] 不存在 visitMap ，更新 visitMap[s[i]] = i

更新 maxLen = max(maxLen, i - start +1)

當 start + maxLen ≥ len(s) 代表已經沒有辦法再找到更長不重複字元的子字串，直接回傳 maxLen

最後回傳 maxLen

import java.util.HashMap;

public class Solution {
  public int lengthOfLongestSubstring(String s) {
    int maxLen = 0, start = 0, sLen = s.length();
    HashMap<Character, Integer> hitMap = new HashMap<>();
    for (int end = 0; end < sLen; end++) {
      char ch = s.charAt(end);
      if (hitMap.containsKey(ch) && hitMap.get(ch) >= start) {
        start = hitMap.get(ch) + 1;
      }
      hitMap.put(ch, end);
      maxLen = Math.max(maxLen, end - start + 1);
      if (start + maxLen >= sLen) {
        break;
      }
    }
    return maxLen;
  }
}

1. 要看出找出不重複字元子字串每個位置間的關係
2. 要理解需要透過 hashTable 去儲存當下遇到每個字元的最新位置，用來做遇到重複字元開始位置更新

• 初始化 start = 0, maxLen = 0, visitMap
• 從 i = 0.. len(s)-1 每次做以下運算
• 檢查 s[i] 是否有在 visitMap 中
• 如果有 , 檢查 visitMap[s[i]] ≥ start ， 更新 start = visitMap[s[i]]+ 1
• 更新 visitMap[s[i]] = i
• 更新 maxLen = max(maxLen, i - start +1)
• 當 start + maxLen ≥ len(s) 代表無法從 start 開始找到比 maxLen 長的不重複字元子字串 直接回傳 maxLen
• 回傳 maxLen