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.
思路一,我自己的思路採用的是hashmap來保存每次出現的字符,當出現重複字符的時候,就記錄這段的最長,然後重新開始。但是這樣很浪費時間,重複了一些步驟。
class Solution {
public int lengthOfLongestSubstring(String s) {
//採用hashmap的簡單解法
int result = 0;
//從第一個字符開始
for(int i = 0; i < s.length(); i++)
{
Map<Character, Integer> hmap = new HashMap<>();
int count = 0;
//計算從第一個字符開始的不重複子串
for(int j = i ; j < s.length(); j++)
{
if(!hmap.containsKey(s.charAt(j)))
{
hmap.put(s.charAt(j),1);
count++;
}
else
break;
}
//保存當前最大的子串
if(count >= result)
result = count;
}
return result;
}
}
思路二:開闢一個能容納所有字母ASCII碼大小的數組,用字母當下標,值當出現的位置,用一個指針來指示從不重複的位置開始。
class Solution {
public int lengthOfLongestSubstring(String s) {
int[] m = new int[256];
Arrays.fill(m, -1);
int res = 0, left = -1;
for (int i = 0; i < s.length(); ++i) {
//當出現重複的時候,從這個重複的位置重新開始,實際上是從下一個位置開始的
left = Math.max(left, m[s.charAt(i)]);
m[s.charAt(i)] = i;//字母當下標,位置當值
//求每次不重複的子串
res = Math.max(res, i - left);
}
return res;
}
}