3. Longest Substring Without Repeating Characters
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.
題解：
輸入一個(gè)字符串，求這個(gè)字符串中，不包括重復(fù)元素的最大的子串的長(zhǎng)度；
例如對(duì)于輸入："zffcool"
子串為：
"z", "zf", "zff", "zffc", "zffco", "zffcoo","zffcool"；
"f", "ff", "ffc", "ffco", "ffcoo","ffcool"；
"f", "fc", "fco", "fcoo","fcool"；
"c", "co", "coo","cool"；
"o", "oo","ool"；
"o", "ol"；
"l"；
則加粗的為不包含重復(fù)字符的子串；
最長(zhǎng)的子串為fco，長(zhǎng)度為3；輸出3；
分析：
兩個(gè)指針，begin 指向子串的頭，i 向后滑動(dòng)；
當(dāng)滑到字符曾出現(xiàn)過時(shí)，判斷重復(fù)的字符是否在子串中，在的話更新 begin，讓它指向重復(fù)字符的下一個(gè)字符，同時(shí)更新重復(fù)字符的位置；否則，在 begin 的前面，說明子串中沒有與當(dāng)前字符重復(fù)的字符，直接更新該重復(fù)字符（key）對(duì)應(yīng)的當(dāng)前位置（value）即可；最后輸出最大的子串長(zhǎng)度；
My Solution(C/C++)
#include <cstdio>
#include <iostream>
#include <string>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
int lengthOfLongestSubstring(string s) {
if (s.length() == 0) {
return 0;
}
map<char, int> hash_map;
vector<int> result;
int count = 0;
int begin = 0;
for (int i = 0; i < s.length(); i++) {
if (hash_map.find(s[i]) == hash_map.end()) {
hash_map.insert(map<char, int>::value_type(s[i], i));
count += 1;
result.push_back(count);
}
else {
if (begin > hash_map[s[i]]) {
count += 1;
}
else {
count = count - (hash_map[s[i]] + 1 - begin) + 1;
begin = hash_map[s[i]] + 1;
}
hash_map[s[i]] = i;
result.push_back(count);
}
}
return *max_element(result.begin(), result.end());
}
};
int main() {
Solution s;
printf("\"abcabcbb\"：%d\n", s.lengthOfLongestSubstring("abcabcbb"));
printf("\"bbbbb\"：%d\n", s.lengthOfLongestSubstring("bbbbb"));
printf("\"pwwkew\"：%d\n", s.lengthOfLongestSubstring("pwwkew"));
printf("\"\"：%d\n", s.lengthOfLongestSubstring(""));
printf("\"nfpdmpi\"：%d\n", s.lengthOfLongestSubstring("nfpdmpi"));
printf("\"nffdmpn\"：%d\n", s.lengthOfLongestSubstring("nffdmpn"));
printf("\"aabaab!bb\"：%d", s.lengthOfLongestSubstring("aabaab!bb"));
return 0;
}
結(jié)果
"abcabcbb"：3
"bbbbb"：1
"pwwkew"：3
""：0
"nfpdmpi"：5
"nffdmpn"：5
"aabaab!bb"：3
My Solution1:
class Solution(object):
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
count, ss = [0], ''
for i in range(len(s)):
if s[i] in ss:
ss = ss[ss.index(s[i])+1: ] + s[i]
count.append(len(ss))
else:
ss += s[i]
if len(ss) > 0:
count.append(len(ss))
return max(count)
My Solution2:
class Solution(object):
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
if s == '':
return 0
begin, long_s, result = 0, '', 0
for end in range(len(s)):
if long_s.find(s[end]) == -1:
long_s += s[end]
if result < len(long_s):
result = len(long_s)
else:
begin = long_s.find(s[end]) + 1
long_s = long_s[begin: ] + s[end]
return result
Reference (轉(zhuǎn)):
class Solution:
# @return an integer
def lengthOfLongestSubstring(self, s):
start = maxLength = 0
usedChar = {}
for i in range(len(s)):
if s[i] in usedChar and start <= usedChar[s[i]]:
start = usedChar[s[i]] + 1
else:
maxLength = max(maxLength, i - start + 1)
usedChar[s[i]] = i
return maxLength
