leetcode - 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 the palindrome partitioning

["aa","b"]

could be produced using 1 cut.

动态规划：

记忆化搜索实现，能pass small data set，但运行large data set里 TLE了。

public class Solution {
private java.util.Map<String, Integer> map = new java.util.HashMap<String, Integer>();
private java.util.Map<String, Boolean> palinMap = new java.util.HashMap<String, Boolean>();
public int minCut(String s) {
if (s.length() == 1)
return 0;

if (map.get(s) != null)
return map.get(s);

if (isPalindrome(s)) {
map.put(s, 0);
return 0;
}

int ret = s.length()-1;

for (int i=1; i<s.length(); i++) {
ret = Math.min(ret, minCut(s.substring(0, i))+1+minCut(s.substring(i)));
}
map.put(s, ret);
return ret;
}

public boolean isPalindrome(String s) {
if (palinMap.get(s) != null)
return palinMap.get(s);
boolean ret = true;
int i=0, j=s.length()-1;
while (i < j) {
char b = s.charAt(i++);
char e = s.charAt(j--);
if (b != e) {
ret = false;
break;
}
}
palinMap.put(s, ret);
return ret;
}
}

这里采用了记忆化搜索方法记录了字符串是否为回文串的信息，但依然TLE，看样子得想别的办法了，该算法改用递推法实现为：

public class Solution {

private java.util.Map<String, Boolean> palinMap = new java.util.HashMap<String, Boolean>();
public int minCut(String s) {
int len = s.length();
int[][] data = new int[len][len];
for (int i=0; i<len; i++)
for (int j=0; j<len; j++)
if (i==j)
data[i][j] = 0;
else
data[i][j] = s.length()-1;
for (int add=1; add<len; add++)
for (int i=0; i<len&&(i+add)<len; i++) {
if (isPalindrome(s.substring(i, i+add+1))) {
data[i][i+add] = 0;
} else {
for (int d=1; d<=add; d++) {
data[i][i+add] = Math.min(data[i][i+add], data[i][i+d-1]+data[i+d][i+add]+1);
}
}
}
return data[0][len-1];
}

public boolean isPalindrome(String s) {
if (palinMap.get(s) != null)
return palinMap.get(s);
boolean ret = true;
int i=0, j=s.length()-1;
while (i < j) {
char b = s.charAt(i++);
char e = s.charAt(j--);
if (b != e) {
ret = false;
break;
}
}
palinMap.put(s, ret);
return ret;
}
}