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Given a string s, partition s such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s.
For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.
思路: 除了用flag[i][j]记录从i到j是否是Palindrome，还要用cut[i]存储到i位置位置最少的cut数
class Solution { public: int minCut(string s) { int sLength = s.length(); vector<vector<bool>> flag(sLength,vector<bool>(sLength, false)); int cut[sLength]; cut[0]=0; for (int i=1;i<sLength;i++) { cut[i]=1+cut[i-1]; for(int j=0;j<i;j++) if(s[i]==s[j]&&(i-j<=2||flag[j+1][i-1])) { flag[j][i]=true; if(j==0) { cut[i] = 0; break; } else cut[i] = min(cut[i],1+cut[j-1]); } } return cut[sLength-1]; } };
132. Palindrome Partitioning II (String; DP)
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原文地址：http://www.cnblogs.com/qionglouyuyu/p/4905676.html
