一、最長無重複子串
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.
分析：本題使用滑動窗口，利用map數據結構，頭指針不停前移，如果頭指針指向的當前字符已經存在於map中，那麼尾指針取當前位置於該字符位置的最大值(尾指針看情況前移)，之後更新ans並將當前字符納入map。
class Solution {
public:
int lengthOfLongestSubstring(string s) {
map<char,int> hash;
int ans=0;
//map.find通過key找value
for(int i=0,j=0;j<s.size();j++)
{
if(hash.find(s[i])!=hash.end())//找到了
{
i = max(hash.find(s[j])->second,i);
}
ans = max(ans,j-i+1);
hash[s[j]] = j+1;
}
return ans;
}
};
二、最長迴文串
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example 1:
Input: "babad" Output: "bab" Note: "aba" is also a valid answer.
Example 2:
Input: "cbbd" Output: "bb"
分析：
1.可使用dp，定義狀態i,j爲字符串i到j是否爲迴文串
P(i,j)=(P(i+1,j−1) and Si==Sj)
The base cases are:
P(i,i)=true P(i, i) = true P(i,i)=true
class Solution {
public:
string longestPalindrome(string s) {
int l = s.size();
if(l==0)
return "";
bool dp[l][l] ;
for(int i=0;i<l;i++)
for(int j=0;j<l;j++)
dp[i][j]=false;
int mi=0, mj=1;
for(int i=0;i<l;i++)
{
dp[i][i] = true;
if(i != l-1 )
if(s[i]==s[i+1])
{
dp[i][i+1]=true;
mi=i;
mj=2;
}
}
for(int d=2;d<l;d++)
for(int i=0;i<l-d;i++)
{
int j=i+d;
if(dp[i+1][j-1] && s[i]==s[j])
{
dp[i][j]=true;
mi=i;
mj=d+1;
}
}
cout<<mi<<mj;
return s.substr(mi,mj);//mi爲起始位置，j爲長度
}
};
2.可以使用從中心擴展法
分析：迴文串最多有2n-1箇中心，我們遍歷這些中心，尋找最長串
class Solution {
public:
string longestPalindrome(string s) {
int l = s.size();
if(l==0)
return "";
int start = 0;
int len;
for(int i=0;i<l-1;i++)
{
int len1 = expend(s,i,i);
int len2 = expend(s,i,i+1);
if(max(len1,len2)>len)
{
len = max(len1,len2);
start = i - (len-1)/2 ;
}
cout<<start<<' '<<len<<endl;
}
return s.substr(start,len);
}
int expend(string s, int L,int R)
{
while(L>=0 && R<s.size() && s[L]==s[R])
{
L--;
R++;
}
return R-L-1;
}
};
