5. Longest Palindromic Substring
2017-12-16 09:24
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1、题目描述
输入一个字符串s,返回最长的回文子串。
2、思路
方法1:
动归。
dp[i][j] refers to the whether the string is palindromic, which if the substring of s starting from the index i and ending at the index j.
dp[i][j] = dp[i+1][j-1] if s[i]==s[j],
otherwise, dp[i][j] = false.
Note: dp[i][j] is always true when i == j or i+1==j and s[i]==s[j].
复杂度 O(n^2)。
方法2:
先从头找出字符串的一小段完全相同的子串(长度可以是1),这个子串一定是回文的,再在它的两头扩展子串,
直至不回文为止,产生的这个新的子串如果更长,则更新答案。
空间复杂度比动归小。
3、代码
方法1:
string longestPalindrome(string s) {
int len = s.size();
bool dp[len][len];
int x=0,y=0;
for(int i=0;i<len;i++)
for(int j=0;j<len;j++)
dp[i][j]=false;
for(int i=len-1;i>=0;i--){
for(int j=i;j<len;j++){
if(i==j)
dp[i][j]=true;
else{
if(s[i]==s[j]){
if(i+1<=j-1)
dp[i][j]=dp[i+1][j-1];
else
dp[i][j]=true;
}
}
if(dp[i][j]){
if(j-i>y-x){
y=j;
x=i;
}
}
}
}
return s.substr(x,y-x+1);
}
方法2:
string longestPalindrome(string s) {
if (s.empty()) return "";
if (s.size() == 1) return s;
int min_start = 0, max_len = 1;
for (int i = 0; i < s.size();) {
if (s.size() - i <= max_len / 2) break;
int j = i, k = i;
while (k < s.size()-1 && s[k+1] == s[k]) ++k; // Skip duplicate characters.
i = k+1;
while (k < s.size()-1 && j > 0 && s[k + 1] == s[j - 1]) { ++k; --j; } // Expand.
int new_len = k - j + 1;
if (new_len > max_len) { min_start = j; max_len = new_len; }
}
return s.substr(min_start, max_len);
}
输入一个字符串s,返回最长的回文子串。
2、思路
方法1:
动归。
dp[i][j] refers to the whether the string is palindromic, which if the substring of s starting from the index i and ending at the index j.
dp[i][j] = dp[i+1][j-1] if s[i]==s[j],
otherwise, dp[i][j] = false.
Note: dp[i][j] is always true when i == j or i+1==j and s[i]==s[j].
复杂度 O(n^2)。
方法2:
先从头找出字符串的一小段完全相同的子串(长度可以是1),这个子串一定是回文的,再在它的两头扩展子串,
直至不回文为止,产生的这个新的子串如果更长,则更新答案。
空间复杂度比动归小。
3、代码
方法1:
string longestPalindrome(string s) {
int len = s.size();
bool dp[len][len];
int x=0,y=0;
for(int i=0;i<len;i++)
for(int j=0;j<len;j++)
dp[i][j]=false;
for(int i=len-1;i>=0;i--){
for(int j=i;j<len;j++){
if(i==j)
dp[i][j]=true;
else{
if(s[i]==s[j]){
if(i+1<=j-1)
dp[i][j]=dp[i+1][j-1];
else
dp[i][j]=true;
}
}
if(dp[i][j]){
if(j-i>y-x){
y=j;
x=i;
}
}
}
}
return s.substr(x,y-x+1);
}
方法2:
string longestPalindrome(string s) {
if (s.empty()) return "";
if (s.size() == 1) return s;
int min_start = 0, max_len = 1;
for (int i = 0; i < s.size();) {
if (s.size() - i <= max_len / 2) break;
int j = i, k = i;
while (k < s.size()-1 && s[k+1] == s[k]) ++k; // Skip duplicate characters.
i = k+1;
while (k < s.size()-1 && j > 0 && s[k + 1] == s[j - 1]) { ++k; --j; } // Expand.
int new_len = k - j + 1;
if (new_len > max_len) { min_start = j; max_len = new_len; }
}
return s.substr(min_start, max_len);
}
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