leetcode: Palindrome Partitioning II

这题在Palindrome Partitioning的基础上还需要一次动态规划。

dp[i][j]保存i到j是否是回文

dp2[i]保存从0到i需要的最小cut数

状态转换方程：dp[i] = min(dp[j])+1 ( 0=< j < i)

class Solution {public:    int minCut(string s) {        if( s.size() <= 1)            return 0;        int size = s.size();        vector< vector< bool> > dp( size, vector< bool>( size, false));        for( int i = 0; i < size; ++i)            dp[i][i] = true;        for( int i = 0; i < size; ++i){            for( int j = 0; j < i; ++j){                if( s[j] == s[i] && dp[j+1][i-1]){                    dp[j][i] = true;                }                if( s[j] == s[i] && j == i - 1){                    dp[j][i] = true;                }            }        }        vector< int> dp2(size, 0);        int tmp = 0;        for( int i = 0; i < size; ++i){            if(dp[0][i]){                dp2[i] = 0;                continue;            }            tmp = 0x7fffffff;            for( int j = 0; j < i; ++j){                if(dp[j+1][i]){                    if(tmp > dp2[j]+1){                        tmp = dp2[j] + 1;                    }                }                dp2[i] = tmp;            }        }        return dp2[size-1];    }};