problem
solution
- string
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22class Solution {
public:
bool isPalindromic(string str)
{
int l=0,r = str.size()-1;
while(l<r)
{
if(str[l++] != str[r--]) return false;
}
return true;
}
int countSubstrings(string s) {
// burse force , bfs
int n = s.size();
int ret = 0;
for(int i=0;i<n;++i)
{
for(int j=i;j<n;++j) ret+=isPalindromic(s.substr(i, j-i+1));
}
return ret;
}
}; - dp
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37class Solution {
public:
int countSubstrings(string s) {
//
// a a a
//a 1 1 1
//a 1 1
//a 1
// b a b a b
//b 1 0 1 0 1
//a 1 0 1 0
//b 1 0 1
//a 1 0
//b 1
int ret = 0;
int n= s.size();
vector<vector<bool>> dp(n, vector<bool>(n, false));
for(int i=0;i<n;++i) dp[i][i] = true;
ret +=n;
for(int i=n-2;i>-1 ; i--){
for(int j=i+1;j<n;++j){
if(s[i] == s[j]) {
if(j==i+1) dp[i][j] = true;
else if(dp[i+1][j-1]) dp[i][j] =true;
}
if(dp[i][j]) ret++;
}
}
return ret;
}
};
analysis
- string
- time complexity
O(n^3) - space complexity
O(1)
- time complexity
- dp
- time complexity
O(nm) - space complexity
O(nm)
- time complexity