資訊專(zhuān)欄INFORMATION COLUMN
摘要：哈希表法雙指針?lè)ㄖ挥挟?dāng)也就是時(shí)上面的循環(huán)才會(huì)結(jié)束，，跳過(guò)這個(gè)之前的重復(fù)字符再將置為
Problem
Given a string, find the length of the longest substring without repeating characters.
ExampleFor example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3.
For "bbbbb" the longest substring is "b", with the length of 1.
Note Solution1. 哈希表法
HashMap Method I
public class Solution {
public int lengthOfLongestSubstring(String s) {
HashMap map = new HashMap();
int count = 0, start = 0;
for (int i = 0; i < s.length(); i++) {
if (map.containsKey(s.charAt(i))) {
int index = map.get(s.charAt(i));
for (int j = start; j <= index; j++) {
map.remove(s.charAt(j));
}
start = index + 1;
}
map.put(s.charAt(i), i);
count = Math.max(count, i - start + 1);
}
return count;
}
}
HashMap Method II
public class Solution {
public int lengthOfLongestSubstring(String s) {
if (s == null || s.length() == 0) return 0;
HashMap map = new HashMap<>();
int max = 1, start = 0;
for (int i = 0; i < s.length(); i++) {
Character cur = s.charAt(i);
Integer rep = map.get(cur);
if (rep != null && rep >= start) {
max = Math.max(max, i-start);
start = rep+1;
}
map.put(cur, i);
}
return Math.max(s.length()-start, max);
}
}
2. 雙指針?lè)?/strong>
public class Solution {
public int lengthOfLongestSubstring(String s) {
int start = 0, end = 0, count = 0;
boolean[] mark = new boolean[256];
while (end < s.length()) {
if (!mark[s.charAt(end)]) {
mark[s.charAt(end)] = true;
count = Math.max(count, end-start+1);
end++;
}
else {
while (mark[s.charAt(end)]) {
mark[s.charAt(start)] = false;
start++;
}
//只有當(dāng)s.charAt(start) == s.charAt(end)
//也就是mark[s.charAt(end)] == false時(shí)
//上面的循環(huán)才會(huì)結(jié)束，start++，跳過(guò)這個(gè)之前的重復(fù)字符
//再將mark[s.charAt(end)]置為true
mark[s.charAt(end)] = true;
end++;
}
}
return count;
}
}
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