Longest Substring Without Repeating Characters
Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: "pwwkew" Output: 3 Explanation: 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：给出一个指定字符串，找出所有字符各不相同，且最长的子串，返回其长度。
这里提供三种解法：
1. 暴力求解, 挨个列出全部子串，逐一比较 时间复杂度 O（n三次方）
2. 使用 HashSet 时间复杂度 O（n）
3. 使用 HashMap 时间复杂度 O（n）
package pers.leetcode;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
/**
* LeetCode 03 难易程度: Medium
*
* @author admin
* @date 2019/3/13 13:01
*/
public class LongestSubString {
public static void main(String[] args) {
String s = "abcabcbb";
int maxSubStinglenggth = slidingWindow(s);
System.out.println("最长子串长度 : " + maxSubStinglenggth);
}
/**
* 暴力求解
*
* @param s 字符串
* @return 子串长度
*/
public static int lengthOfLongestSubString(String s){
int n = s.length();
int ans = 0;
for (int i = 0; i < n; i++){
for (int j = i; j <= n; j++){
if (allUnique(s, i, j)){
ans = Math.max(ans, j-i);
}
}
}
return ans;
}
/**
* 判断子串是否字符是否各不相同
*
* @param s 字符串
* @param start 起始位置
* @param end 终结位置
* @return 是否每个都是唯一的
*/
public static boolean allUnique(String s, int start, int end){
Set<Character> set = new HashSet<>();
for (int i = start; i < end; i++){
Character ch = s.charAt(i);
if (set.contains(ch)){
return false;
}
set.add(ch);
}
return true;
}
/**
* 解法二 : 滑动窗口
*
* @param s 字符串
*/
public static int slidingWindow(String s){
int n = s.length();
Set<Character> set = new HashSet<>();
int ans = 0, i = 0, j = 0;
while (i < n && j < n){
//尝试扩展[i,j]的范围
if (!set.contains(s.charAt(j))){
set.add(s.charAt(j++));
ans = Math.max(ans, j - i);
}
else {
set.remove(s.charAt(i++));
}
}
return ans;
}
/**
* 最优化解法, 使用 HashMap
*
* @param s 字符串
*/
public static int optimized(String s){
int n = s.length(), ans = 0;
// current index of character
Map<Character, Integer> map = new HashMap<>();
// try to extend the range [i, j]
for (int j = 0, i = 0; j < n; j++) {
if (map.containsKey(s.charAt(j))) {
i = Math.max(map.get(s.charAt(j)), i);
}
ans = Math.max(ans, j - i + 1);
map.put(s.charAt(j), j + 1);
}
return ans;
}
}