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 <= 10000
S only contains characters "I" or "D".
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<Int>(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<String>) {
val input = "IIID"
val diStringMatch = DIStringMatch()
CommonUtils.printArray(diStringMatch.diStringMatch(input).toTypedArray())
}
有问题随时沟通
