LeetCode-in-All

131. Palindrome Partitioning

Medium

Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.

A palindrome string is a string that reads the same backward as forward.

Example 1:

Input: s = “aab”

Output: [[“a”,”a”,”b”],[“aa”,”b”]]

Example 2:

Input: s = “a”

Output: [[“a”]]

Constraints:

• 1 <= s.length <= 16
• s contains only lowercase English letters.

Solution

object Solution {
    def partition(s: String): List[List[String]] = {
        val res = scala.collection.mutable.ListBuffer[List[String]]()
        backtracking(res, List.empty[String], s, 0)
        res.toList
    }

    private def backtracking(res: scala.collection.mutable.ListBuffer[List[String]], currArr: List[String], s: String, start: Int): Unit = {
        if (start == s.length) {
            res += currArr.reverse
        }
        for (end <- start until s.length) {
            if (isPalindrome(s, start, end)) {
                backtracking(res, s.substring(start, end + 1) :: currArr, s, end + 1)
            }
        }
    }

    private def isPalindrome(s: String, start: Int, end: Int): Boolean = {
        var left = start
        var right = end
        while (left < right && s(left) == s(right)) {
            left += 1
            right -= 1
        }
        left >= right
    }
}

This site is open source. Improve this page.