维护一个窗口,每次关注窗口中的字符串,在每次判断中,左窗口和右窗口选择其一向前移动。维护一个HashSet, 正常情况下移动右窗口,如果没有出现重复则继续移动右窗口,如果发现重复字符,则说明当前窗口中的串已经不满足要求,继续移动有窗口不可能得到更好的结果,此时移动左窗口,直到不再有重复字符为止,中间跳过的这些串中不会有更好的结果,因为他们不是重复就是更短。因为左窗口和右窗口都只向前,所以两个窗口都对每个元素访问不超过一遍,因此时间复杂度为O(2*n)=O(n),是线性算法。空间复杂度为HashSet的size,也是O(n). 代码如下:
package edu.bupt.cici.leetcode; import java.util.ArrayList; import java.util.HashMap; import java.util.HashSet; import java.util.Iterator; public class LongestSubstringWithoutRepeatingCharacters { public int lengthOfLongestSubstring(String s) { if (s == null && s.length() == 0) return 0; HashSet<Character> set = new HashSet<Character>(); int max = 0; int walker = 0; int runner = 0; while (runner < s.length()) { if (set.contains(s.charAt(runner))) { if (max < runner - walker) { max = runner - walker; } while (s.charAt(walker) != s.charAt(runner)) { set.remove(s.charAt(walker)); walker++; } walker++; } else { set.add(s.charAt(runner)); } runner++; } max = Math.max(max, runner - walker); return max; } public static void main(String[] args) { // TODO Auto-generated method stub LongestSubstringWithoutRepeatingCharacters sol = new LongestSubstringWithoutRepeatingCharacters(); String s = "abcaaaaabcd"; System.out.println(sol.lengthOfLongestSubstring(s)); } }
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