Palindrome Partitioning II

Q:

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.

A:

跟I类似，还是先保存每个子字符串是否为回文。同样A[i]与A[0].....A[i - 1]息息相关，还是比较easy的

class Solution {
public:
    int minCut(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        Prepare(s);
        if (s.empty()) return 0;
        int len = s.length();
        min_cuts_.resize(len, 0);
        for (int i = 1; i < len; ++i) {
            if (is_pal_[0][i]) {
                min_cuts_[i] = 0;
                continue;
            }
            min_cuts_[i] = i;
            for (int j = 0; j < i; ++j) {
                if (!is_pal_[j + 1][i]) continue;
                min_cuts_[i] = min(min_cuts_[j] + 1, min_cuts_[i]);
            }
        }
        return min_cuts_.back();
    }
private:
    
    void Prepare(const string& s) {
        min_cuts_.clear();
        is_pal_.clear();
        if (s.empty()) return;
        int len = s.length();
        is_pal_.resize(len, vector<bool>());
        for (int i = 0; i < len; ++i) {
            is_pal_[i].resize(len, false);
        }
        for (int i = 0; i < len; ++i) is_pal_[i][i] = true;
        for (int i = len - 2; i >= 0; --i) {
            is_pal_[i][i + 1] = (s[i] == s[i + 1]);
            for (int j = i + 2; j < len; ++j) {
                is_pal_[i][j] = (s[i] == s[j] && is_pal_[i + 1][j - 1]);
            }
        }
    }
    
    vector<int> min_cuts_;
    vector<vector<bool> > is_pal_;
};