Leet Code OJ 3. Longest Substring Without Repeating Characters

题目

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given “abcabcbb”, the answer is “abc”, which the length is 3.

Given “bbbbb”, the answer is “b”, with the length of 1.

Given “pwwkew”, 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.

翻译

给定一个字符串，找到最长的子串的长度，这个子串不存在重复的字符。
例如：
输入”abcabcbb”, 符合条件的子串就是”abc”, 长度为3。

输入”bbbbb”, 符合条件的子串就是”b”, 长度为1

输入”pwwkew”, 符合条件的子串就是”wke”, 长度为3。提示：要求必须是子串,例如”pwke” 是一个子序列但不是一个子串。

分析

遍历一遍，记录符合条件的长度count。如果遇到相同的字符，则从左往右remove直到移除掉那个相同的字符。

Java版代码（时间复杂度O(n),空间复杂度O(n)）:

public class Solution {public int lengthOfLongestSubstring(String s) {Map<Character, Integer> map = new HashMap<Character, Integer>();char[] charArr = s.toCharArray();int max = 0;int count = 0;for (int i = 0; i < charArr.length; i++) {Integer last = map.get(charArr[i]);if (last == null) {count++;if (count > max) {max = count;}} else {int start=i-count;for(int j=start;j<last;j++){map.remove(charArr[j]);count--;}}map.put(charArr[i], i);}return max;}
}

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Leet Code OJ 3. Longest Substring Without Repeating Characters

题目

翻译

分析

Java版代码（时间复杂度O(n),空间复杂度O(n)）:

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