update 132

Author: selfboot

DynamicProgramming/132_PalindromePartitioningII.cpp

@@ -0,0 +1,60 @@
+/*
+ * @Author: xuezaigds@gmail.com
+ * @Last Modified time: 2016-10-11 10:06:28
+ */
+
+class Solution {
+public:
+    /*
+    Dynamic Programming:
+    cuts[i]: minimum cuts needed for a palindrome partitioning of s[i:], so we want cuts[0].
+
+    To get cuts[i-1], we scan j from i-1 to len(s)-1.
+    Once we comes to a is_palindrome[i-1][j]==true:
+        if j==len(s)-1, the string s[i-1:] is a Pal, cuts[i-1] is 0;
+        else: the current cut num (first cut s[i-1:j+1] and then cut the rest
+        s[j+1:]) is 1+cuts[j+1], compare it to the exisiting cuts[i-1], repalce if smaller.
+
+    is_palindrome[i][j]: whether s[i:j+1] is palindrome.
+    Here we need not to compute the is_palindrome in advance.
+    We use "Dynamic Programming" too, the formula is very intuitive:
+    is_palindrome[i][j] = true if (is_palindrome[i+1][j-1] and s[i] == s[j]) else false
+
+    A better O(n) space solution can be found here:
+    https://discuss.leetcode.com/topic/2840/my-solution-does-not-need-a-table-for-palindrome-is-it-right-it-uses-only-o-n-space
+    */
+
+    int minCut(string s) {
+        if(s=="")   return 0;
+        int len = s.size();
+        std::vector<int> cuts(len, 0);
+        // Initialize the cuts array.
+        for(int i=0; i<len; i++){
+            cuts[i] = len - i-1;
+        }
+        std::vector<std::vector<bool>> is_palindrome(len, std::vector<bool>(len, false));
+        for(int i=len-1; i>=0; i--){
+            for(int j=i; j<len; j++){
+                if((j-i<=2 && s[i]==s[j]) || (s[i]==s[j] && is_palindrome[i+1][j-1])){
+                    is_palindrome[i][j]=true;
+                    if(j==len-1){
+                        cuts[i]=0;
+                    }
+                    else{
+                        cuts[i] = min(cuts[i], cuts[j+1]+1);
+                    }
+                }
+            }
+        }
+        return cuts[0];
+    }
+};
+
+/*
+""
+"aab"
+"aabb"
+"aabaa"
+"acbca"
+"acbbca"
+*/

DynamicProgramming/132_PalindromePartitioningII.py

@@ -5,9 +5,23 @@
 class Solution(object):
     """
     Dynamic Programming:
-    cuts[i]: minimum cuts needed for a palindrome partitioning of s[i:]
-    is_palindrome[i][j]: whether s[i:i+1] is palindrome
+    cuts[i]: minimum cuts needed for a palindrome partitioning of s[i:], so we want cuts[0].
+
+    To get cuts[i-1], we scan j from i-1 to len(s)-1.
+    Once we comes to a is_palindrome[i-1][j]==true:
+        if j==len(s)-1, the string s[i-1:] is a Pal, cuts[i-1] is 0;
+        else: the current cut num (first cut s[i-1:j+1] and then cut the rest
+        s[j+1:]) is 1+cuts[j+1], compare it to the exisiting cuts[i-1], repalce if smaller.
+
+    is_palindrome[i][j]: whether s[i:j+1] is palindrome.
+    Here we need not to compute the is_palindrome in advance.
+    We use "Dynamic Programming" too, the formula is very intuitive:
+    is_palindrome[i][j] = true if (is_palindrome[i+1][j-1] and s[i] == s[j]) else false
+
+    A better O(n) space solution can be found here:
+    https://discuss.leetcode.com/topic/2840/my-solution-does-not-need-a-table-for-palindrome-is-it-right-it-uses-only-o-n-space
     """
+
     def minCut(self, s):
         if not s:
             return 0
@@ -16,28 +30,27 @@ def minCut(self, s):
         is_palindrome = [[False for i in range(s_len)]
                          for j in range(s_len)]
 
-        cuts = [s_len-1-i for i in range(s_len)]
-        for i in range(s_len-1, -1, -1):
+        cuts = [s_len - 1 - i for i in range(s_len)]
+        for i in range(s_len - 1, -1, -1):
             for j in range(i, s_len):
                 # if self.is_palindrome(i, j):
-                if ((j-i < 2 and s[i] == s[j]) or
-                        (s[i] == s[j] and is_palindrome[i+1][j-1])):
+                if ((j - i < 2 and s[i] == s[j]) or (s[i] == s[j] and is_palindrome[i + 1][j - 1])):
                     is_palindrome[i][j] = True
                     if j == s_len - 1:
                         cuts[i] = 0
                     else:
-                        cuts[i] = min(cuts[i], 1+cuts[j+1])
+                        cuts[i] = min(cuts[i], 1 + cuts[j + 1])
                 else:
                     pass
 
         return cuts[0]
 
+
 """
-if __name__ == "__main__":
-    sol = Solution()
-    print sol.minCut("aab")
-    print sol.minCut("aabb")
-    print sol.minCut("aabaa")
-    print sol.minCut("acbca")
-    print sol.minCut("acbbca")
+""
+"aab"
+"aabb"
+"aabaa"
+"acbca"
+"acbbca"
 """

DynamicProgramming/README.md

@@ -103,7 +103,7 @@
 # 优化
 
 115.Distinct Subsequences  
-
+132 
 
 # 更多阅读
 [fibonacci数列为什么那么重要](https://www.zhihu.com/question/28062458)