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Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
For example, given s = "aab",
Return
[
["aa","b"],
["a","a","b"]
]
解题思路一：
将s分为左右两个部分，分别求出两边的partition，然后粘一块即可，JAVA实现如下：
static public List<List<String>> partition(String s) {
Set<List<String>> set = new HashSet<List<String>>();
if (isPalindrome(s)) {
List<String> alist = new ArrayList<String>();
alist.add(s);
set.add(alist);
}
for (int i = 1; i < s.length(); i++) {
List<List<String>> left = partition(s.substring(0, i));
List<List<String>> right = partition(s.substring(i, s.length()));
for (List<String> aLeft : left)
for (List<String> aRight : right) {
List<String> alist = new ArrayList<String>(aLeft);
alist.addAll(aRight);
set.add(new ArrayList<String>(alist));
}
}
return new ArrayList<List<String>>(set);
}
static boolean isPalindrome(String s) {
int left = 0;
int right = s.length() - 1;
while (left < right)
if (s.charAt(left++) != s.charAt(right--))
return false;
return true;
}
结果TLE
解题思路二：
修改下思路一，从左边入手，如果左边是Palindrome，对右边求一个partition，这样求得的结果也不会重复，这样就可以AC了，JAVA实现如下：
static public List<List<String>> partition(String s) {
ArrayList<List<String>> list = new ArrayList<List<String>>();
if (isPalindrome(s)) {
List<String> alist = new ArrayList<String>();
alist.add(s);
list.add(alist);
}
for (int i = 1; i < s.length(); i++)
if (isPalindrome(s.substring(0, i))) {
List<String> aLeft = new ArrayList<String>();
aLeft.add(s.substring(0, i));
List<List<String>> right = partition(s.substring(i, s.length()));
for (List<String> aRight : right) {
List<String> alist = new ArrayList<String>(aLeft);
alist.addAll(aRight);
list.add(new ArrayList<String>(alist));
}
}
return list;
}
static boolean isPalindrome(String s) {
int left = 0;
int right = s.length() - 1;
while (left < right)
if (s.charAt(left++) != s.charAt(right--))
return false;
return true;
}
Java for LeetCode 131 Palindrome Partitioning
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原文地址：http://www.cnblogs.com/tonyluis/p/4544612.html
