标签：style blog color 使用 io for ar art
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：先使用DP确定s中任意两个字符之间是否构成回文。若isPalindrome[step][i]为true，则表示str.substr(i,step+i)构成回文。
然后使用一个两层的循环查找出最小划分数。
1 class Solution { 2 public: 3 int minCut( string s ) { 4 int size = s.length(); 5 if( size <= 1 ) { return 0; } 6 vector<bool> *isPalindrome = new vector<bool>[size]; 7 isPalindrome[0].resize( size, true ); 8 for( int step = 1; step < size; ++step ) { 9 isPalindrome[step].resize( size-step, false ); 10 if( step == 1 ) { 11 for( int i = 0; i < size-1; ++i ) { 12 if( s[i] == s[i+1] ) { isPalindrome[1][i] = true; } 13 } 14 } else { 15 for( int i = 0; i < size-step; ++i ) { 16 if( s[i] == s[i+step] && isPalindrome[step-2][i+1] ) { isPalindrome[step][i] = true; } 17 } 18 } 19 } 20 vector<int> Q( size, 0 ); 21 for( int i = 1; i < size; ++i ) { 22 if( isPalindrome[i][0] ) { continue; } 23 Q[i] = Q[i-1]+1; 24 for( int j = i-1; j > 0; --j ) { 25 if( isPalindrome[i-j][j] && Q[i] > Q[j-1]+1 ) { 26 Q[i] = Q[j-1]+1; 27 if( Q[i] == 1 ) { break; } 28 } 29 } 30 } 31 delete [] isPalindrome; 32 return Q.back(); 33 } 34 };
Palindrome Partitioning II,布布扣,bubuko.com
标签：style blog color 使用 io for ar art
原文地址：http://www.cnblogs.com/moderate-fish/p/3908624.html
