릿코드 1578. Minimum Time to Make Rope Colorful 코틀린

컬러 하나 복제해둔다음 
컬러로 반복문돌리고
현재위치에서 다른색만날때까지 반복문돌리고
1개빼고 다 제거후 제일 작은수하나말고 나머지 제거시간 res에 추가
제거한건 -1로 바꾸고 -1만나면 컨티뉴


타임아웃

class Solution {
	fun minCost(colors: String, neededTime: IntArray): Int {
		var tempColor=colors

		var res=0
		for (i in colors.indices){
			if (tempColor[i]=='0'){
				continue
			}
			val tempTimeList= mutableListOf(neededTime[i])
			for (j in i+1 until tempColor.length){
				if (tempColor[i]!=tempColor[j]){
					break
				}
				tempTimeList.add(neededTime[j])
				tempColor=tempColor.replaceRange(j,j+1,"0")
			}
			tempTimeList.remove(tempTimeList.max())
			println(tempTimeList)
			res+=tempTimeList.sum()

		}
		println(tempColor)
		println(res)

		return res
	}
}

그냥 
현재추가스택배열 만들어두고
for 한번만 돌려서 이전값 하나 기억해두고,달라질때마다 현재추가스택배열에서 제일큰거하나빼고 나머지 res에 추가

정답

class Solution {
    fun minCost(colors: String, neededTime: IntArray): Int {
        var tempColor=colors

        var res=0
        var prevValue='0'
        val tempTimeList= mutableListOf(neededTime[0])
        for (i in colors.indices){
            if (prevValue==colors[i]){
                tempTimeList.add(neededTime[i])
            }else{
                tempTimeList.remove(tempTimeList.max())
                res+=tempTimeList.sum()
                prevValue=colors[i]
                tempTimeList.clear()
                tempTimeList.add(neededTime[i])
            }

        }
        tempTimeList.remove(tempTimeList.max())
        res+=tempTimeList.sum()
        println(tempTimeList)
        println(res)

        return res
    }
}

이중반복문을 돌릴려는 못된버릇을 고치자