*[topcoder]PalindromicSubstringsDiv2

http://community.topcoder.com/stat?c=problem_statement&pm=12967

计算一个字符串里Palindrome的数量。我的DP方法需要n^2的空间。

#include <vector>

#include <string>

using namespace std;



class PalindromicSubstringsDiv2 {

public:

	int count(vector <string> S1, vector <string> S2) {

		string s;

		for (int i = 0; i < S1.size(); i++) {

			s += S1[i];

		}

		for (int i = 0; i < S2.size(); i++) {

			s += S2[i];

		}

		int ret = 0;

		vector<vector<bool>> dp;

		int N = s.size();

		dp.resize(N);

		for (int i = 0; i < N; i++) {

			dp[i].resize(N + 1);

		}

		for (int len = 1; len <= N; len++) {

			for (int i = 0; i + len <= N; i++) { // start pos

				if (len == 1) {

					dp[i][len] = true;

				} else if (len == 2){

					dp[i][len] = (s[i] == s[i + len - 1]);

				} else {

					dp[i][len] = dp[i + 1][len - 2] && (s[i] == s[i + len - 1]);

				}

				ret += (dp[i][len] ? 1 : 0);

			}

		}

		return ret;

	}

};

如果从中间开始往两边扩，就不需要额外空间了~

#include <vector>

#include <string>

using namespace std;



class PalindromicSubstringsDiv2 {

public:

	int count(vector <string> S1, vector <string> S2) {

		string s;

		for (int i = 0; i < S1.size(); i++) {

			s += S1[i];

		}

		for (int i = 0; i < S2.size(); i++) {

			s += S2[i];

		}

		int count = 0;

		int N = s.size();

		for (int m = 0; m < N; m++) {

			for (int even = 0; even < 2; even++) {

				int i, j = 0;

				if (even == 0) {

					i = j = m;

				} else {

					i = m;

					j = m + 1;				

				}

				for (; i >= 0 && j < N; i--,j++) {

					if (s[i] == s[j])

						count++;

					else

						break;

				}

			}

		}

		return count;

	}

};