*[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;
	}
};