方法1: backtracking。这道题最一开始的想法就是backtracking,但是会tle,经过我小小的优化以后只有一个test case tle了,而且我的算法不能避免duplicate,这些都是问题。虽然没有accepted,但是还是展示一下,至少逻辑是没问题的。
class Solution {
public List<String> generatePalindromes(String s) {
Set<String> res = new HashSet<>();
StringBuilder path = new StringBuilder();
backtracking(s, 0, path, res);
return new ArrayList<>(res);
}
public void backtracking(String s, int start, StringBuilder path, Set<String> res){
if(start == s.length()){
StringBuilder ans = new StringBuilder(path);
if(isPalindrome(ans.toString())) res.add(ans.toString());
return;
}
boolean flag = false;
for(int i = 0; i <= path.length(); i++){
if(flag) continue;
if(i != path.length() && s.charAt(start) == path.charAt(i)) flag = true;
path.insert(i, s.charAt(start));
backtracking(s, start+1, path, res);
path.deleteCharAt(i);
}
}
public boolean isPalindrome(String s){
if(s.length() == 0 ||s.length() == 1) return true;
int left = 0;
int right = s.length()-1;
while(left < right){
if(s.charAt(left) != s.charAt(right)) return false;
left++;
right--;
}
return true;
}
}
方法2: 先写一个函数判断这个string能不能构成回文数,如果不行的直接返回空。然后后面还有一个比较tricky的优化是,因为是可以构成回文的,所以我们只选用一半的元素,排列组合出这一半的元素所有可能的permutation,然后后半段直接对称复制上去就行了。时间复杂(n/2)!,空间复杂n/2。详细方法请见lc官方解答2.
public class Solution {
Set < String > set = new HashSet < > ();
public List < String > generatePalindromes(String s) {
int[] map = new int[128];
char[] st = new char[s.length() / 2];
if (!canPermutePalindrome(s, map))
return new ArrayList < > ();
char ch = 0;
int k = 0;
for (int i = 0; i < map.length; i++) {
if (map[i] % 2 == 1)
ch = (char) i;
for (int j = 0; j < map[i] / 2; j++) {
st[k++] = (char) i;
}
}
permute(st, 0, ch);
return new ArrayList < String > (set);
}
public boolean canPermutePalindrome(String s, int[] map) {
int count = 0;
for (int i = 0; i < s.length(); i++) {
map[s.charAt(i)]++;
if (map[s.charAt(i)] % 2 == 0)
count--;
else
count++;
}
return count <= 1;
}
public void swap(char[] s, int i, int j) {
char temp = s[i];
s[i] = s[j];
s[j] = temp;
}
void permute(char[] s, int l, char ch) {
if (l == s.length) {
set.add(new String(s) + (ch == 0 ? "" : ch) + new StringBuffer(new String(s)).reverse());
} else {
for (int i = l; i < s.length; i++) {
if (s[l] != s[i] || l == i) {
swap(s, l, i);
permute(s, l + 1, ch);
swap(s, l, i);
}
}
}
}
}
总结:
- 无