中心展开

Author: yuzhangcmu

algorithm/dp/LongestPalindrome_dp1.java

@@ -0,0 +1,100 @@
+package Algorithms.algorithm.dp;
+
+public class LongestPalindrome_dp1 {
+    public static void main(String[] args) {
+        String s = "cabaabad";
+        System.out.println(longestPalindrome(s));
+    }
+    
+    // Solution 1:
+    public static String longestPalindrome1(String s) {
+        if (s == null) {
+            return null;
+        }
+        
+        int len = s.length();
+
+        // Record i-j is a palindrome.
+        boolean[][] D = new int[len][len];
+        for (int i = 0; i < len; i++) {
+        	for (int j = 0; j < len; j++) {
+        		D[i][j] = false;
+        	}
+        }
+
+        int max = 0;
+        int retB = 0;
+        int retE = 0;
+        // 这样写的目的是，从前往后扫描时，被记录的DP值可以被复用
+        // 因为D[i][j] 要用到i + 1, j - 1，所以每一次计算j时，把j对应的i全部计算完，这样
+        // 下一次计算i,j的时候，可以有i+1, j-1可以用。
+        for (int j = 0; j < len; j++) {
+        	for (int i = 0; i <= j; i++) {
+        	    if (s.charAt(i) == s.charAt(j)
+                    && (j - i <= 3 || D[i + 1][j - 1])
+                    ) {
+                    D[i][j] = true;
+
+                    if (j - i + 1 > max) {
+                        retB = i;
+                        retE = j;    
+                    }
+                } else {
+                    D[i][j] = false;
+                }            
+        	}
+        }
+        
+        return s.substring(retB, retE + 1);
+    }
+
+    // solution 2: 中心展开法。从头扫到尾部，每一个字符以它为中心向两边扩展，找最长回文。
+    // 复杂度为N^2 并且是inplace，空间复杂度O(1)
+    public static String longestPalindrome(String s) {
+        if (s == null) {
+            return null;
+        }
+        
+        int len = s.length();
+
+        if (len <= 0) {
+            return "";
+        }
+
+        int max = 0;
+        String ret = "";
+
+        for (int i = 0; i < len; i++) {
+            // 考虑奇数字符串
+            String s1 = expandAround(s, i, i);
+            if (s1.length() > max) {
+                ret = s1;
+                max = s1.length();
+            }
+
+            String s2 = expandAround(s, i, i);
+            if (s2.length() > max) {
+                ret = s2;
+                max = s2.length();
+            }
+        }
+        
+        return ret;
+    }
+
+    public static String expandAround(String s, int c1, int c2) {
+        int len = s.length();
+        
+        while (c1 >= 0 && c2 <= len - 1) {
+            if (s.charAt(c1) != s.charAt(c2)) {
+                break;    
+            }
+
+            c1++;
+            c2--;
+        }
+        
+        // 注意，根据 substring的定义，c2不要减1
+        return s.substring(c1 + 1, c2);
+    }
+}

dp/LongestPalindrome_dp1.java

@@ -6,8 +6,8 @@ public static void main(String[] args) {
         System.out.println(longestPalindrome(s));
     }
 
-    // Solution 1:
-    public static String longestPalindrome(String s) {
+    // solution 1: DP.
+    public static String longestPalindrome1(String s) {
         if (s == null) {
             return null;
         }
@@ -43,4 +43,55 @@ public static String longestPalindrome(String s) {
 
         return s.substring(retB, retE + 1);
     }
+    
+    // solution 2: 中心展开法。从头扫到尾部，每一个字符以它为中心向两边扩展，找最长回文。
+    // 复杂度为N^2 并且是inplace，空间复杂度O(1)
+    public static String longestPalindrome(String s) {
+        if (s == null) {
+            return null;
+        }
+        
+        int len = s.length();
+
+        if (len <= 0) {
+            return "";
+        }
+
+        int max = 0;
+        String ret = "";
+
+        for (int i = 0; i < len; i++) {
+            // 考虑奇数字符串
+            String s1 = expandAround(s, i, i);
+            if (s1.length() > max) {
+                ret = s1;
+                max = s1.length();
+            }
+
+            // 考虑偶数长度的字符串
+            String s2 = expandAround(s, i, i + 1);
+            if (s2.length() > max) {
+                ret = s2;
+                max = s2.length();
+            }
+        }
+        
+        return ret;
+    }
+
+    public static String expandAround(String s, int c1, int c2) {
+        int len = s.length();
+        
+        while (c1 >= 0 && c2 <= len - 1) {
+            if (s.charAt(c1) != s.charAt(c2)) {
+                break;    
+            }
+
+            c1--;
+            c2++;
+        }
+        
+        // 注意，根据 substring的定义，c2不要减1
+        return s.substring(c1 + 1, c2);
+    }
 }