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
1 class Solution { 2 public: 3 int minCut(string s) { 4 int N = s.size(); 5 bool isP[N]; 6 int dp[N]; 7 dp[0] = 0; 8 for(int i = 1; i < N; ++i) { 9 isP[i] = true; 10 dp[i] = dp[i-1] + 1; 11 for(int j = 0; j < i; ++j) { 12 isP[j] = (s[i] == s[j]) ? isP[j+1] : false; // isP[j] == true -> [j...i] is a palindrome 13 // isP[j+1] == true -> [j+1...i-1] is a palindrome 14 if (isP[j]) 15 dp[i] = (j == 0) ? 0 : min(dp[i], dp[j-1] + 1); // dp[i] -> minCount for [0...i] 16 } 17 } 18 return dp[N-1]; 19 } 20 };
the palindrome partitioning ["aa","b"] could be produced using 1 cut.
Solution: dp. Contributed by 孙冕. Great job:)