更新 test.java

src/solution/test.java

@@ -1,49 +1,39 @@
 package solution;
 
 
+import java.util.HashSet;
+import java.util.Set;
+
 public class test {
     public static void main(String[] args) {
-        String s = "abcabcbb";
-        int length = lengthOfLongestSubstring(s);
-        System.out.println(length);
+        String s = "bbbbb";
+        String s1 = "abcabcbb";
+        String s2 = "pwwkew";
+        String s3 = "ckilbkd";
+        String s4 = "dvdf";
+        System.out.println(lengthOfLongestSubstring(s));
 
     }
-
     public static int lengthOfLongestSubstring(String s) {
-        StringBuilder longestString = new StringBuilder();
-        StringBuilder lastMaxString = new StringBuilder();
-        if (s != null){
-            for (int i = 0; i < s.length(); i++) {
-                char temp = s.charAt(i);
-                if (i == 0){
-                    longestString.append(temp);
-                    continue;
-                }
-                Character prChar = s.charAt(i-1);
-                if (prChar != temp){
-                    boolean in = isInString(temp,longestString);
-                    if (!in){
-                        longestString.append(temp);
-                    }
-
-                } else {
-                    lastMaxString = longestString;
-                }
+        // 哈希集合，记录每个字符是否出现过
+        Set<Character> occ = new HashSet<Character>();
+        int n = s.length();
+        // 右指针，初始值为 -1，相当于我们在字符串的左边界的左侧，还没有开始移动
+        int rk = -1, ans = 0;
+        for (int i = 0; i < n; ++i) {
+            if (i != 0) {
+                // 左指针向右移动一格，移除一个字符
+                occ.remove(s.charAt(i - 1));
             }
-            return longestString.length();
+            while (rk + 1 < n && !occ.contains(s.charAt(rk + 1))) {
+                // 不断地移动右指针
+                occ.add(s.charAt(rk + 1));
+                ++rk;
+            }
+            // 第 i 到 rk 个字符是一个极长的无重复字符子串
+            ans = Math.max(ans, rk - i + 1);
         }
-        return 0;
+        return ans;
     }
 
-    public static boolean isInString(Character a,StringBuilder b){
-        if (b != null) {
-            for (int i = 0; i < b.length(); i++) {
-                Character temp = b.charAt(i);
-                if (temp == a) {
-                    return true;
-                }
-            }
-        }
-        return false;
-    }
 }