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Given a string, find the length of the longest substring without repeating characters.
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.
/***暴力破解法* 时间复杂度O(N*N*N)* @param* @param start* @param end* @return*/public static boolean allUnique(String s, int start, int end){Set<Character> set = new HashSet<Character>();for(int i = start; i < end; i++){if(set.contains(s.charAt(i))) return false;set.add(s.charAt(i));}return true;}public static int lengthOfLongestSubstring1(String s) {int len = s.length();int ans = 0;for(int i = 0; i < len; i++){for(int j = i+1; j<=len;j++){if(allUnique(s,i,j)){ans = Math.max(ans,j-i);}}}return ans;}/*** 滑动窗口法* 时间发杂度O(N*N)* @param s* @return*/public static int lengthOfLongestSubstring2(String s)
{Set<Character> set = new HashSet<Character>();int len = s.length();int i = 0,j=0;int ans = 0;while(i<len && j<len){if(!set.contains(s.charAt(j))){ans = Math.max(ans,j-i+1);set.add(s.charAt(j++));}else{//将已经添加到set集合中的元素从集合中删除，直到集合中不再包含第j个元素为止set.remove(s.charAt(i++));}}return ans;
}/*** 滑动窗口优化* @param s* @return*/public static int lengthOfLongestSubstring3(String s)
{int len = s.length();int ans = 0;Map<Character,Integer>map = new HashMap<>();int i = 0;for(int j = 0; j < len; j++){if(map.containsKey(s.charAt(j))){//获取到当前元素在map中的value,可以理解为索引值，即为无重复字符的长度i = Math.max(map.get(s.charAt(j)),i);}ans = Math.max(ans, j-i+1);map.put(s.charAt(j),j+1);}return ans;
}
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