19. Palindrome Partitioning && Palindrome Partitioning II （回文分割）

Palindrome Partitioning

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab", Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

思想： 简单的深度优先搜索。

bool isPalindrome(string& s, int l, int r) {
    while(l++ < r--) 
        if(s[l] != s[r]) return false;
    return true;
}
class Solution {
public:
    void dfs(string& s, vector<string>& vec2, size_t id) {
        if(id == s.size()) {
            vec.push_back(vec2);
            return;
        }
        for(int end = id; end < s.size(); ++end) {
            if(isPalindrome(s, id, end)) {
                vec2.push_back(s.substr(id, end-id+1));
                dfs(s, vec2, end+1);
                vec2.pop_back();
            }
        }
    }
    vector<vector<string> > partition(string s) {
        if(s == "") return vec;
        vector<string> vec2;
        dfs(s, vec2, 0);
        return vec;
    }
private:
    vector<vector<string> > vec;
};

 

Palindrome Partitioning II

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.

思想： 动态规划:

n = s.length();

Record[i] =                                 0                                          , ( i = n || isPalindrome(i, n-1))

                   min(n-1-i, Record[k]+1 ( isPalindrome(i, k) ) )        , otherwise

where i belong to interval [0, n].

class Solution {
public:
    int minCut(string s) {
        if(s == "" || s.size() == 1) return 0;
        int n = s.size();
        vector<vector<bool> > D(n, vector<bool>(n, false));

        vector<int> record(n, 0);
        for(int i = n-1; i >= 0; --i) {
            record[i] = n-1-i;
            for(int j = i; j < n; ++j) {
				if(s[i] == s[j] && (j-i < 2 || D[i+1][j-1])) {
					D[i][j] = true;
					if(j == n-1) record[i] = 0;
					else record[i] = min(record[i], record[j+1]+1);
				}
            }
        }
        return record[0];
    }
};