[Leetcode] 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.

动态规划，dp[i]表示从s_0到s_i的最小cut，为了避免重复比较，定义标记矩阵flag[i][j]，用来存储s_i到s_j是否为回文。

 1 class Solution {

 2 public:

 3     int minCut(string s) {

 4         vector<int> dp(s.length(), 0);

 5         vector<vector<bool> > flag(s.length(), vector<bool>(s.length(), false));

 6         flag[0][0] = true;

 7         for (int i = 1; i < s.length(); ++i) {

 8             flag[i][i] = true;

 9             dp[i] = dp[i-1] + 1;

10             for (int len = 1; len < i + 1; ++len) {

11                 if (s[i-len] == s[i] && (len < 2 || flag[i-len+1][i-1])) {

12                     dp[i] = (i - len == 0) ? 0 : min(dp[i], dp[i-len-1] + 1);

13                     flag[i-len][i] = true;

14                 }

15             }

16         }

17         return dp[s.length() - 1];

18     }

19 };

换个字母代码更好理解一些。

 1 class Solution {

 2 public:

 3     int minCut(string s) {

 4         vector<int> dp(s.length(), 0);

 5         vector<vector<bool> > flag(s.length(), vector<bool>(s.length(), false));

 6         flag[0][0] = true;

 7         for (int j = 1; j < s.length(); ++j) {

 8             flag[j][j] = true;

 9             dp[j] = dp[j-1] + 1;

10             for (int i = 0; i < j; ++i) {

11                 if (s[i] == s[j] && (i == j - 1 || flag[i+1][j-1])) {

12                     dp[j] = (i == 0) ? 0 : min(dp[j], dp[i-1] + 1);

13                     flag[i][j] = true;

14                 }

15             }

16         }

17         return dp[s.length()-1];

18     }

19 };

 

DFS版本超时了。

 1 class Solution {

 2 public:

 3     bool isPail(string &s) {

 4         int low = 0, high = s.length() - 1;

 5         while (low < high) {

 6             if (s[low] != s[high]) {

 7                 return false;

 8             }

 9             ++low;

10             --high;

11         }

12         return true;

13     }

14     

15     void findNext(vector<vector<string> > &res, string &s, vector<string> part, int idx) {

16         if (idx == s.length()) {

17             res.push_back(part);

18             return;

19         }

20         string substr;

21         for (int len = s.length() - idx; len > 0; --len) {

22             substr = s.substr(idx, len);

23             if (isPail(substr)) {

24                 part.push_back(substr);

25                 findNext(res, s, part, idx + len);

26                 part.pop_back();

27             }

28         }

29     }

30     

31     int minCut(string s) {

32         vector<vector<string> > res;

33         vector<string> part;

34         findNext(res, s, part, 0);

35         int min = INT_MAX;

36         for (int i = 0; i < res.size(); ++i) {

37             min = (min > res[i].size()) ? res[i].size() : min;

38         }

39         return min - 1;

40     }

41 };