标签:Partitioning Palindrome return cur ArrayList start 131 palindrome backtracking
package LeetCode_131 /** * 131. Palindrome Partitioning * https://leetcode.com/problems/palindrome-partitioning/description/ * * Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. Example: Input: "aab" Output: [ ["aa","b"], ["a","a","b"] ] * */ class Solution { private val result = ArrayList<ArrayList<String>>() /* * backtracking, Time complexity:O(2^n), Space complexity:O(n) * */ fun partition(s: String): List<List<String>> { backtracking(0, s, ArrayList()) return result } private fun backtracking(start: Int, s: String, cur: ArrayList<String>) { val n = s.length if (start == n) { val temp = ArrayList<String>(cur) result.add(temp) return } for (i in start until n) { //if s[start,i] is not palindrome, just go to s[start,i+1] if (!isPalindrome(s, start, i)) { continue } cur.add(s.substring(start,i+1)) backtracking(i + 1, s, cur) cur.removeAt(cur.size - 1) } } private fun isPalindrome(s: String, i_: Int, j_: Int): Boolean { var i = i_ var j = j_ while (i < j) { if (s[i] != s[j]) { return false } i++ j-- } return true } }
标签:Partitioning,Palindrome,return,cur,ArrayList,start,131,palindrome,backtracking 来源: https://www.cnblogs.com/johnnyzhao/p/13020167.html
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