릿코드 1903. Largest Odd Number in String 코틀린

앞에서부터 temp에 더하고 홀수면 res덮어쓰기

타임아웃(이지인데 10^5네)

class Solution {
    fun largestOddNumber(num: String): String {
        var res=""
        var temp=""
        for (i in num){
            temp+=i
            if (Character.getNumericValue(i)%2==1){
                res=temp
            }
        }
        println(res)
        return res
    }
}

뒤에서부터 홀수찾은다음 그위치까지 서브스트링

정답

class Solution {
    fun largestOddNumber(num: String): String {
        var res=""
        var count=-1
        for ((index,i) in num.reversed().withIndex()){
            if (Character.getNumericValue(i)%2==1){
                count=index
                println("""index:${index} i:${i}""")
                break
            }
        }

        if (count==-1){
            return ""
        }
        println("""count:${count}""")
        res=num.substring(0,num.length-count)
        println(res)
        return res
    }
}