[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 };

 

posted @ 2014-04-11 01:24  Eason Liu  阅读(508)  评论(0)  编辑  收藏  举报