Java Leetcode Longest Substring Without Repeating Characters
2016-11-09 09:23
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Java Leetcode Longest Substring Without Repeating Characters详解
本文是实现关于leetcode Longest Substring Without Repeating Characters的实现;题目的地址:
https://leetcode.com/problems/longest-substring-without-repeating-characters/
题目的意思:求最长的无重复字符的子串。
写完以后搜了下网上其他的解法,发现有很多时间复杂度o(n3)的解法,这种解法通过暴力遍历来获取结果,在写的时候没有想过这种解法,这种写法估计我也写不出来。
该篇主要讲解的时间复杂度是o(n);是通过容器和临时容器来实现的,容器是存储当前最优的结果,临时容器在遍历的过程中在一直改变,下面代码详细的实现了该算法。PS:在类的注释上用到了窗口的概念是网上看到别人这样用词,代码的实现绝对原创。
具体实现:
/** * 用HashMap容器来装载数据,便于判断是否是重复 first:最终结果的开始位置 end:最终结果的结束位置 length:最大长度 * tempFirst:临时容器开始位置 tempEnd:临时容器结束位置 tempLength:临时容器的大小。 * PS:这里的容器是类似于窗口的概念,窗口看到的是tempFirst,tempEnd之间的内容 * 详细:按照窗口的移动寻找最长的字段。 每遍历一个字符, * 如果该字符没有重复的,则窗口的tempEnd值往后面移动, 如果该字符是重复的,则判断是不是第一次出现的重复字符, * 如果是第一次出现的重复字符,给first,end,length赋值。 * 如果不是第一次出现的字符,并且,当前窗口的大小大于已经找到的窗口的大小,此时需要给first,end,length赋值, * 并且窗口的其实位置要在前一个其实位置和valueInt + 1中取最大值,否则在abba这种情况下,得出的结构是3而不是2 * PS:这里的first和end,仅仅是便于获取满足条件的字符串 */ public class Solution { public int lengthOfLongestSubstring(String s) { if (s.length() == 0 || s.length() == 1) { return s.length(); } HashMap<String, Integer> map = new HashMap<String, Integer>(); int first = 0, end = -1, length = -1; int tempFirst = 0, tempEnd = 0, tempLength = 0; for (int i = 0; i < s.length(); ++i) { if (map.containsKey(String.valueOf(s.charAt(i)))) { if (length == -1 || (length != -1 && tempLength > length)) { first = tempFirst; end = tempEnd; length = tempLength; } int valueInt = map.get(String.valueOf(s.charAt(i))); tempFirst = tempFirst > (valueInt + 1) ? tempFirst : (valueInt + 1); } tempEnd = i; tempLength = tempEnd - tempFirst + 1; map.put(String.valueOf(s.charAt(i)), i); } return tempLength > length ? tempLength : length; } }
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