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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]; }};
19. Palindrome Partitioning && Palindrome Partitioning II (回文分割)