double DP

Author: zhongn2

Palindrome Partitioning II ,cpp.h

@@ -0,0 +1,32 @@
+
+//state: F[i] from 0 - i position the # of minimum cut
+//function: f[i] = min(f[j] + 1 j < i && s[i , j] is palindrome)
+// initialization: f[0] = 0
+// answer: f[s.size()]
+//here is another problem, check a string is palindrome is also time-comsuming
+// so we will use another DP to solve that
+// s[i][j] = s[i-1][j-1] i < j && s[i] == s[j]
+// this problem turns out to be a double DP problem
+class Solution {
+public:
+    int minCut(string s) {
+       const int n = s.size();
+       int f[n + 1];
+       bool p[n][n];
+       fill_n(&p[0][0], n * n, false);
+       for(int i = 0; i <= n; i++){
+           f[i] = n - 1 - i; // worst case each char is a cut
+       }
+       for(int i = n - 1; i >= 0; i--){
+           for(int j = i; j < n; j++){
+               if(s[i] == s[j] && (j -i < 2 ||p[i+1][j-1])){
+                   p[i][j] = true;
+                   f[i] = min(f[i], f[j+1] + 1);
+               }
+           }
+       }
+       return f[0];
+     
+    }
+      
+};

Roman to Integer.cpp

@@ -27,4 +27,5 @@ class Solution {
         }
         return result;
     }
-};
+};
+