3. Longest Substring Without Repeating Characters(c++) 15ms
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.
思路1:既然是找没有重复的字符串,那么重复字符出现的地方就很重要了,于是使用容器放不重复的字符串,遇到重复的字符就清空重复字符串前的字符。执行时间29ms.
class Solution {
public:
int lengthOfLongestSubstring(string s) {
vector<char> vc;
int maxlen = ;
int n = s.size();
for (int i = ; i < n; i++)
{
int cc = s[i];
for (int j = ; j < vc.size(); j++)
{
if (vc[j] == cc)
{
if (maxlen < vc.size())
{
maxlen = vc.size();
}
vc.erase(vc.begin(), vc.begin()+j+);
break;
}
}
vc.push_back(cc);
}
if (maxlen < vc.size())
{
maxlen = vc.size();
}
return maxlen;
}
};
思路2:这个问题实际上是一个动态规划问题,动态规划是通过拆分问题,定义问题状态和状态之间的关系,使得问题能够以递推(或者说分治)的方式去解决。比如这一题,要找没有重复的最长字符串,首先考虑已经找到了第n个字符时最大字符串长度为L,
即S(n)=L,那么遍历第n+1个字符时,如果这个字符已经在前面重复了,可知S(n+1)=L,否则S(n+1)=L+1. 此方法时间复杂度O(n),执行时间15ms.
class Solution {
public:
int lengthOfLongestSubstring(string s) {
vector<int> vc(,-);//使用vector来记录出现的字符
int start=-,maxLen=;
for(int i=;i!=s.size();i++)
{
if(vc[s[i]]>start)
start = vc[s[i]];
vc[s[i]]=i;
maxLen = max(maxLen,i-start);
}
return maxLen;
}
};
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