릿코드 79. Word Search 코틀린

연결된 해당 단어가 배열에 있는지 확인
length가 작아서 시간복잡도 신경안써도됨
그냥 첫글자 찾고,위아래왼오로 다음거있나 재귀하고,만약 그게 이미 갔던자리면 패스하는식으로 하면될듯



정답

class Solution {
    fun exist(board: Array<CharArray>, word: String): Boolean {
        val firstWord = word[0]
        for ((iIndex,iValue) in board.withIndex()){
            for ((jIndex,jValue) in iValue.withIndex()){
                val wordFind = wordFind(board, Pair(iIndex, jIndex), listOf(), word, 0)
                if (wordFind){
                    return true
                }

            }
        }
        return false
    }
    fun wordFind(board: Array<CharArray>, point:Pair<Int,Int>,prevList:List<Pair<Int,Int>>,
                 word: String,wordIndex:Int ):Boolean{
        if (point.first>=board.size || point.second>=board[0].size || point.first<0 ||point.second<0){
            return false
        }
        if (prevList.contains(point)){
            return false
        }
        if (board[point.first][point.second]!=word[wordIndex]){
            return false
        }
        if (word.length-1==wordIndex){
            return true
        }
        val prevCurList = prevList.toMutableList()
        prevCurList.add(point)
        var res=wordFind(board,Pair(point.first+1,point.second),prevCurList,word,wordIndex+1)
        res=res.or(wordFind(board,Pair(point.first-1,point.second),prevCurList,word,wordIndex+1))
        res=res.or(wordFind(board,Pair(point.first,point.second+1),prevCurList,word,wordIndex+1))
        res=res.or(wordFind(board,Pair(point.first,point.second-1),prevCurList,word,wordIndex+1))

        return res
    }
}