132.Palindrome Partitioning II
class Solution {
public:
int minCut(string s) {
if (s.empty()) return 0;
int n = s.size();
vector<vector<bool>> p(n, vector<bool>(n));
vector<int> dp(n);
for (int i = 0; i < n; ++i) {
dp[i] = i;
for (int j = 0; j <= i; ++j) {
if (s[i] == s[j] && (i - j < 2 || p[j + 1][i - 1])) {
p[j][i] = true;
dp[i] = (j == 0) ? 0 : min(dp[i], dp[j - 1] + 1);
}
}
}
return dp[n - 1];
}
};

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