40 lines
1.2 KiB
Java
40 lines
1.2 KiB
Java
package solution;
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import java.util.HashSet;
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import java.util.Set;
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public class test {
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public static void main(String[] args) {
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String s = "bbbbb";
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String s1 = "abcabcbb";
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String s2 = "pwwkew";
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String s3 = "ckilbkd";
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String s4 = "dvdf";
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System.out.println(lengthOfLongestSubstring(s3));
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}
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public static int lengthOfLongestSubstring(String s) {
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// 哈希集合，记录每个字符是否出现过
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Set<Character> occ = new HashSet<Character>();
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int n = s.length();
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// 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
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int rk = -1, ans = 0;
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for (int i = 0; i < n; ++i) {
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if (i != 0) {
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// 左指针向右移动一格，移除一个字符
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occ.remove(s.charAt(i - 1));
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}
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while (rk + 1 < n && !occ.contains(s.charAt(rk + 1))) {
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// 不断地移动右指针
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occ.add(s.charAt(rk + 1));
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++rk;
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}
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// 第 i 到 rk 个字符是一个极长的无重复字符子串
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ans = Math.max(ans, rk - i + 1);
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}
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return ans;
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}
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}
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