add 131, 132

Backtracking/131_PalindromePartitioning.py

@@ -0,0 +1,51 @@
+#! /usr/bin/env python
+# -*- coding: utf-8 -*-
+
+
+class Solution(object):
+    def partition(self, s):
+        if not s:
+            return []
+        self.result = []
+        self.end = len(s)
+        self.str = s
+
+        self.is_palindrome = [[False for i in range(self.end)]
+                              for j in range(self.end)]
+
+        for i in range(self.end-1, -1, -1):
+            for j in range(self.end):
+                if i > j:
+                    pass
+                elif j-i < 2 and s[i] == s[j]:
+                    self.is_palindrome[i][j] = True
+                elif self.is_palindrome[i+1][j-1] and s[i] == s[j]:
+                    self.is_palindrome[i][j] = True
+                else:
+                    self.is_palindrome[i][j] = False
+
+        self.palindrome_partition(0, [])
+        return self.result
+
+    def palindrome_partition(self, start, sub_strs):
+        if start == self.end:
+            # It's confused the following sentence doesn't work.
+            # self.result.append(sub_strs)
+            self.result.append(sub_strs[:])
+            return
+
+        for i in range(start, self.end):
+            if self.is_palindrome[start][i]:
+                sub_strs.append(self.str[start:i+1])
+                self.palindrome_partition(i+1, sub_strs)
+                sub_strs.pop()      # Backtracking here
+
+
+if __name__ == "__main__":
+    sol = Solution()
+    print sol.partition("aab")
+    print sol.partition("aabb")
+    print sol.partition("aabaa")
+    print sol.partition("acbca")
+    print sol.partition("acbbca")
+

DepthFirstSearch/README.md

@@ -0,0 +1,8 @@
+深度优先搜索策略
+
+
+
+
+
+注意复位当前节点属性。
+

DynamicProgramming/132_PalindromePartitioningII.py

@@ -0,0 +1,43 @@
+#! /usr/bin/env python
+# -*- coding: utf-8 -*-
+
+
+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
+    """
+    def minCut(self, s):
+        if not s:
+            return 0
+        s_len = len(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):
+            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])):
+                    is_palindrome[i][j] = True
+                    if j == s_len - 1:
+                        cuts[i] = 0
+                    else:
+                        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")
+"""

README.md

@@ -92,6 +92,7 @@
 * 052. N-Queens II
 * 079. Word Search
 * 093. [Restore IP Addresses](Backtracking/93_RestoreIPAddresses.py)
+* 131. [Palindrome Partitioning](Backtracking/131_PalindromePartitioning.py)
 
 # [Recursion](Recursion/)
 
@@ -116,6 +117,7 @@
 * 120. [Triangle](DynamicProgramming/120_Triangle.py)
 * 121. [Best Time to Buy and Sell Stock](DynamicProgramming/121_BestTimeToBuyAndSellStock.py)
 * 123. [Best Time to Buy and Sell Stock III](DynamicProgramming/123_BestTimeToBuyAndSellStockIII.py)
+* 132. [Palindrome Partitioning II](DynamicProgramming/132_PalindromePartitioningII.py)
 
 # [Greedy](Greedy/)
 

ToBeOptimized/131_PalindromePartitioning.py

@@ -0,0 +1,51 @@
+#! /usr/bin/env python
+# -*- coding: utf-8 -*-
+
+
+class Solution(object):
+    def partition(self, s):
+        if not s:
+            return []
+        self.result = []
+        self.end = len(s)
+        self.str = s
+
+        self.is_palindrome = [[False for i in range(self.end)]
+                              for j in range(self.end)]
+
+        for i in range(self.end-1, -1, -1):
+            for j in range(self.end):
+                if i > j:
+                    pass
+                elif j-i < 2 and s[i] == s[j]:
+                    self.is_palindrome[i][j] = True
+                elif self.is_palindrome[i+1][j-1] and s[i] == s[j]:
+                    self.is_palindrome[i][j] = True
+                else:
+                    self.is_palindrome[i][j] = False
+
+        self.palindrome_partition(0, [])
+        return self.result
+
+    def palindrome_partition(self, start, sub_strs):
+        if start == self.end:
+            # It's confused the following sentence doesn't work.
+            # self.result.append(sub_strs)
+            self.result.append(sub_strs[:])
+            return
+
+        for i in range(start, self.end):
+            if self.is_palindrome[start][i]:
+                sub_strs.append(self.str[start:i+1])
+                self.palindrome_partition(i+1, sub_strs)
+                sub_strs.pop()      # Backtracking here
+
+
+if __name__ == "__main__":
+    sol = Solution()
+    print sol.partition("aab")
+    print sol.partition("aabb")
+    print sol.partition("aabaa")
+    print sol.partition("acbca")
+    print sol.partition("acbbca")
+

ToBeOptimized/132_PalindromePartitioningII.py

@@ -0,0 +1,43 @@
+#! /usr/bin/env python
+# -*- coding: utf-8 -*-
+
+
+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
+    """
+    def minCut(self, s):
+        if not s:
+            return 0
+        s_len = len(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):
+            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])):
+                    is_palindrome[i][j] = True
+                    if j == s_len - 1:
+                        cuts[i] = 0
+                    else:
+                        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")
+"""

ToBeOptimized/132_PalindromePartitioningII_Optimized.py

@@ -0,0 +1,44 @@
+#! /usr/bin/env python
+# -*- coding: utf-8 -*-
+# Refer to: https://leetcode.com/discuss/9476/solution-does-not-need-table-palindrome-right-uses-only-space
+# A better solution
+
+
+class Solution(object):
+    """
+    Dynamic Programming:
+    """
+    def minCut(self, s):
+        s_len = len(s)
+        # number of minnum cuts for the pre i characters
+        min_cuts = [i-1 for i in range(s_len+1)]
+
+        for i in range(s_len):
+            # odd length palindrome
+            j = 0
+            while i-j >= 0 and i+j < s_len:
+                if s[i-j] == s[i+j]:
+                    min_cuts[i+j+1] = min(min_cuts[i+j+1], min_cuts[i-j]+1)
+                    j += 1
+                else:
+                    break
+            # even length palindrome
+            j = 1
+            while i-j+1 >= 0 and i+j < s_len:
+                if s[i-j+1] == s[i+j]:
+                    min_cuts[i+j+1] = min(min_cuts[i+j+1], min_cuts[i-j+1]+1)
+                    j += 1
+                else:
+                    break
+
+        return min_cuts[s_len]
+
+"""
+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")
+"""