Title description
Give you a string s, please split s into some substrings so that each substring is a palindrome string. Return all possible splitting schemes for s.
A palindrome is a string that is read both forward and backward.
Example 1:
Input: s = "aab"
Output: [["a","a","b"],["aa","b"]]
Example 2:
Input: s = "a"
Output: [["a"]]
Problem-solving ideas
-
Each node represents the remaining strings that have not been scanned, and the branch generated is to intercept the prefix of the remaining strings;
-
When generating the prefix string, determine whether the prefix string is a palindrome.
- If the prefix string is a palindrome, branches and nodes can be generated;
- If the prefix string is not a palindrome, no branches and nodes will be generated. This step is a pruning operation.
-
Settlement when the leaf node is an empty string. At this time, the path from the root node to the leaf node is a result in the result set. Use depth-first traversal to record all possible results.
-
Use a path variable path to search, and use one path globally (note that a copy is generated during settlement), so after the recursive execution method ends, it needs to backtrack, that is, take out the elements added before the recursion;
-
The path operation is only at the end of the list, so the appropriate data structure is the stack.
class Solution_131 {
public List<List<String>> partition(String s) {
int len = s.length();
List<List<String>> res = new ArrayList<>();
if (len == 0){
return res;
}
Deque<String> stack = new ArrayDeque<>();
char[] charArray = s.toCharArray();
dfs(charArray,0,len,stack,res);
return res;
}
/**
*
* @param charArray
* @param index 起始字符的索引
* @param len 字符串s的长度，可以设置为全局变量
* @param path 记录从根节点到叶子节点的路径
* @param res 记录所有的结果
*/
private void dfs(char[] charArray, int index, int len, Deque<String> path, List<List<String>> res) {
if (index == len){
res.add(new ArrayList<>(path));
return;
}
for (int i = index; i <len; i++){
//因为截取字符串是消耗性能的，采用传子串的方式判断一个子串是否是回文子串
if (!checkPalindrome(charArray,index,i)){
continue;
}
path.addLast(new String(charArray,index,i+1-index));
dfs(charArray,i+1,len,path,res);
path.removeLast();
}
}
/**
* 采用动态规划，把回文子串的结果记录在一个表格里
* @param charArray
* @param left 子串的左边界，可以取到
* @param right 子串的右边界，可以取到
* @return
*/
private boolean checkPalindrome(char[] charArray, int left, int right) {
while (left < right){
if (charArray[left] != charArray[right]){
return false;
}
left++;
right--;
}
return true;
}
}