题目链接：
https://leetcode-cn.com/problems/palindromic-substrings/
难度：中等
647. 回文子串
给定一个字符串，你的任务是计算这个字符串中有多少个回文子串。
具有不同开始位置或结束位置的子串，即使是由相同的字符组成，也会被视作不同的子串。
示例 1：
输入："abc"
输出：3
解释：三个回文子串: "a", "b", "c"
示例 2：
输入："aaa"
输出：6
解释：6个回文子串: "a", "a", "a", "aa", "aa", "aaa"
提示：
输入的字符串长度不会超过 1000 。
How do you say this question? After all, it is still a spicy chicken.
The simplest idea is of course to brute force to enumerate all strings and then determine whether it is a palindrome. The code is as follows
class Solution {
public:
int countSubstrings(string s) {
int n = s.size();
int ans=0;
for(int i=0;i<n;++i){
for(int j=i;j<n;++j){
if(is(s,i,j)){
ans++;
}
}
}
return ans;
}
bool is(string &s,int i,int j){
while(i<=j){
if(s[i]!=s[j]){
return false;
}
i++;
j--;
}
return true;
}
};
Although the way of thinking is extremely simple, the time complexity is O(n 3 ) and the execution time is very long. . . Then I looked at the official problem solution and I knew that I decided to have a better
solution. The idea of this solution is also very simple (I didn't think of it before, but I heard it for the first time). Enumerate the center of all strings and expand from the center (of course, two pointers are needed)
But here is a problem. The center of the parity string is different. When it is an odd string, its center is one character. When it is an even string, its center is two strings
. There are two solutions: one is the odd-even case. Cite (I didn’t try this
one ) The other is: List the number (the number refers to the number that starts from 0) and the enumeration center instance to find the law: left=i/2 (rounded down) right=left+i %2 also find out the relationship between the length of the string and the maximum number (that is, how many possible situations are there in total) =2*n-1
（n=4）
编号 left right
0 0 0
1 0 1
2 1 1
3 1 2
4 2 2
5 2 3
6 3 3
class Solution {
public:
int countSubstrings(string s) {
int n = s.size();
int ans=0;
for(int i=0;i<2*n-1;++i){
int left=i/2;
int right=left+i%2;
while(left>=0&&right<n){
if(s[left]==s[right]){
left--;
right++;
ans++;
}else{
break;
}
}
}
return ans;
}
};
There is another way. . . I heard it for the first time: I learned about the Manacher algorithm online and didn't figure out the principle. Although the code is written, it is written according to the algorithm. . . Record a coordinate
Manacher algorithm
algorithm problem solution
class Solution {
public:
int countSubstrings(string s) {
int ans=0;
string t="$#";
for(const char& c:s){
t+=c;
t+='#';
}
t+='!';
int n = t.size()-1;
vector<int> flag(n);
int imax=0;
int rmax=0;
for(int i=1;i<n;++i){
if(i<=rmax){
flag[i]=min(flag[2*imax-i],rmax-i+1);
}else{
flag[i]=1;
}
while(t[i+flag[i]]==t[i-flag[i]]){
flag[i]++;
}
if(flag[i]-1+i>rmax){
imax=i;
rmax=flag[i]-1+i;
}
ans+=(flag[i]/2);
}
return ans;
}
};