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LeetCode #Longest Substring Without Repeating Characters#
折腾了一回，第一感觉就是没有感觉．．．这种没技巧的话暴力是不现实的．．而后发现其实没有很好理解题目的意
思考过么？究竟什么是最长的子字符串
这里有个很要命的概念， 子字符串，别小看这家伙．是搞定题目的关键．
"abcduiwe" 这串字符串是没有重复字符的，那么最长子串就是自己．
但是！一旦出现重复的字符了，当前最大子字符串必然变小．
比方说，原字符串为＂abcd＂，最大子字符串就是自身．
但这个时候加上了一个字符b
"abcdb"那么这个时候b就作为子串abcd的结束位置了！这就让我们的最大子串（自身），变小了
变成了去除b的前面部分
再看，如果我们加的不同的字符更多
"abcdbefgh"这个时候，最长的子串就是"befgh", 要点就在于，从重复的那个字符串开始算新的子字符串！
刚回来的路上还一直闷神，，＂子字符串．．子字符串＂
"""
Programmer : EOF
e-mail : jasonleaster@gmail
Date : 2015.04.02
File : lswrc.py
"""
class Solution:
# @return an integer
def lengthOfLongestSubstring(self, s):
Table = [-1 for i in range(0, 256)]
maxLen = 0
lastRepeatPos = -1
"""
We will use @ord() function to translate the character
into the number of it's ascii code.
"""
for i in range(0, len(s)):
if Table[ord(s[i])] != -1 and lastRepeatPos < Table[ord(s[i])]:
lastRepeatPos = Table[ord(s[i])]
if i - lastRepeatPos > maxLen :
maxLen = i - lastRepeatPos
Table[ord(s[i])] = i
return maxLen
#---------- just for testing ----------
s = Solution()
print s.lengthOfLongestSubstring("c")
print s.lengthOfLongestSubstring("abcdababcde")
下面是 @凯旋冲锋 的java的实现
"""
Programmer : EOF
e-mail : jasonleaster@gmail
Date : 2015.04.02
File : lswrc.py
"""
class Solution:
# @return an integer
def lengthOfLongestSubstring(self, s):
Table = [-1 for i in range(0, 256)]
maxLen = 0
lastRepeatPos = -1
"""
We will use @ord() function to translate the character
into the number of it's ascii code.
"""
for i in range(0, len(s)):
if Table[ord(s[i])] != -1 and lastRepeatPos < Table[ord(s[i])]:
lastRepeatPos = Table[ord(s[i])]
if i - lastRepeatPos > maxLen :
maxLen = i - lastRepeatPos
Table[ord(s[i])] = i
return maxLen
#---------- just for testing ----------
s = Solution()
print s.lengthOfLongestSubstring("c")
print s.lengthOfLongestSubstring("abcdababcde")
皓神的C++实现：
// Source : https://oj.leetcode.com/problems/longest-substring-without-repeating-characters/
// Author : Hao Chen
// Date : 2014-07-19
/**********************************************************************************
*
* Given a string, find the length of the longest substring without repeating characters.
* For 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.
*
**********************************************************************************/
#include <string.h>
#include <iostream>
#include <string>
#include <map>
using namespace std;
/*
* Idea:
*
* Using a map store each char's index.
*
* So, we can be easy to know the when duplication and the previous duplicated char's index.
*
* Then we can take out the previous duplicated char, and keep tracking the maxiumn length.
*
*/
int lengthOfLongestSubstring1(string s) {
map<char, int> m;
int maxLen = 0;
int lastRepeatPos = -1;
for(int i=0; i<s.size(); i++){
if (m.find(s[i])!=m.end() && lastRepeatPos < m[s[i]]) {
lastRepeatPos = m[s[i]];
}
if ( i - lastRepeatPos > maxLen ){
maxLen = i - lastRepeatPos;
}
m[s[i]] = i;
}
return maxLen;
}
//don't use <map>
int lengthOfLongestSubstring(string s) {
const int MAX_CHARS = 256;
int m[MAX_CHARS];
memset(m, -1, sizeof(m));
int maxLen = 0;
int lastRepeatPos = -1;
for(int i=0; i<s.size(); i++){
if (m[s[i]]!=-1 && lastRepeatPos < m[s[i]]) {
lastRepeatPos = m[s[i]];
}
if ( i - lastRepeatPos > maxLen ){
maxLen = i - lastRepeatPos;
}
m[s[i]] = i;
}
return maxLen;
}
int main(int argc, char** argv)
{
string s = "abcabcbb";
cout << s << " : " << lengthOfLongestSubstring(s) << endl;
s = "bbbbb";
cout << s << " : " << lengthOfLongestSubstring(s) << endl;
s = "bbabcdb";
cout << s << " : " << lengthOfLongestSubstring(s) << endl;
if (argc>1){
s = argv[1];
cout << s << " : " << lengthOfLongestSubstring(s) << endl;
}
return 0;
}
LeetCode #Longest Substring Without Repeating Characters#
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原文地址：http://blog.csdn.net/cinmyheart/article/details/44837895
