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];
}
};