LeetCode之DI String Match（Kotlin）-白红宇

问题： Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length. Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1:

If S[i] == "I", then A[i] < A[i+1]

If S[i] == "D", then A[i] > A[i+1]

Example 1:Input: "IDID"Output: [0,4,1,3,2]Example 2:Input: "III"Output: [0,1,2,3]Example 3:Input: "DDI"Output: [3,2,0,1] Note:1 <= S.length <= 10000S only contains characters "I" or "D".复制代码

方法：如果是增加就是最小数，如果是减少就是最大数；最大数后就是次大数，最小数之后就是次小数。当然这只是所有情况中的一种，也是最容易代码化的。

具体实现：

class DIStringMatch {    fun diStringMatch(S: String): IntArray {        var ins = S.length        var des = 0        val result = arrayOfNulls
    
     (S.length + 1)        var index = 0        for (ch in S) {            if(ch == 'I') {                result[index] = des                des++            } else {                result[index] = ins                ins--            }            index++        }        // result[index] = ins        result[index] = des        return result.requireNoNulls().toIntArray()    }}fun main(args: Array
     
      ) {    val input = "IIID"    val diStringMatch = DIStringMatch()    CommonUtils.printArray(diStringMatch.diStringMatch(input).toTypedArray())}复制代码

有问题随时沟通