Title: Given "abc", all possible segmentation combinations are given, and all elements in the combination must be palindrome strings.
example1: [a, b, c], [a, bc], [ab, c], [abc], the result is only [a, b, c] is a palindrome example2: give aab: all cut
combinations Form [a,a,b], [aa,b], [a,ab], [aab], only [a,a,b], [aa,b] are palindrome strings
/**
* Author：m
* Date： 2023/04/11 22:51
*/
public class SplitAbcResultPalindrome {
public static void main(String[] args) {
String candidate = "aac";
List<List<String>> res = dfs(candidate);
System.out.println(res);
}
private static List<List<String>> dfs(String candidate) {
List<List<String>> result = Lists.newArrayList();
if (StringUtils.isBlank(candidate)) {
return result;
}
List<String> combine = Lists.newArrayList();
int startIndex = 0;
recursion(result, combine, candidate, startIndex);
return result;
}
private static void recursion(List<List<String>> result, List<String> combine, String candidate, int startIndex) {
// 1.终止条件
// 2.小集合添加至大集合
if (startIndex == candidate.length()) {
result.add(Lists.newArrayList(combine));
}
// 3.循环
for (int i = startIndex; i < candidate.length(); i++) {
String subStr = candidate.substring(startIndex, i + 1);
// 3.1 去重|过滤(这里要求元素必须为回文串)
if (!strIsPalindrome(subStr)) {
continue;
}
// 3.2优化
// 3.3添加元素至小集合
combine.add(subStr);
// 3.4递归(i+1)
recursion(result, combine, candidate, i + 1);
// 3.5回溯(删除小集合中最后一个元素)
combine.remove(combine.size() - 1);
}
}
private static boolean strIsPalindrome(String str) {
if (StringUtils.isBlank(str)) {
return false;
}
if (str.length() == 1) {
return true;
}
char[] chars = str.toCharArray();
for (int i = 0, j = chars.length - 1; i < j; i++, j--) {
if (chars[i] != chars[j]) {
return false;
}
}
return true;
}
}