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  • *[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;
    	}
    };
    

      

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  • 原文地址:https://www.cnblogs.com/lautsie/p/3901593.html
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