LeetCode之Delete Columns to Make Sorted（Kotlin）

問題： We are given an array A of N lowercase letter strings, all of the same length. Now, we may choose any set of deletion indices, and for each string, we delete all the characters in those indices. The remaining rows of strings form columns when read north to south. For example, if we have an array A = ["abcdef","uvwxyz"] and deletion indices {0, 2, 3}, then the final array after deletions is ["bef","vyz"], and the remaining columns of A are ["b","v"], ["e","y"], and ["f","z"]. (Formally, the c-th column is [A[0][c], A[1][c], ..., A[A.length-1][c]].) Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order. Return the minimum possible value of D.length.

Example 1:

Input: ["cba","daf","ghi"]
Output: 1
Explanation: 
After choosing D = {1}, each column ["c","d","g"] and ["a","f","i"] are in non-decreasing sorted order.
If we chose D = {}, then a column ["b","a","h"] would not be in non-decreasing sorted order.
Example 2:

Input: ["a","b"]
Output: 0
Explanation: D = {}
Example 3:

Input: ["zyx","wvu","tsr"]
Output: 3
Explanation: D = {0, 1, 2}
 

Note:

1 <= A.length <= 100
1 <= A[i].length <= 1000
複製代碼

---

方法： 按列去遍歷，找出不是正常字母順序的列數即爲結果。

具體實現：

class DeleteColumnsToMakeSorted {
    fun minDeletionSize(A: Array<String>): Int {
        var lastChar: Char
        var result = 0
        for (j in A[0].indices) {
            lastChar = 'a'
            for (i in A.indices) {
                if (A[i][j] < lastChar) {
                    result++
                    break
                }
                lastChar = A[i][j]
            }
        }
        return result
    }
}

fun main(args: Array<String>) {
    val input = arrayOf("zyx","wvu","tsr")
    val deleteColumnsToMakeSorted = DeleteColumnsToMakeSorted()
    println(deleteColumnsToMakeSorted.minDeletionSize(input))
}
複製代碼

有問題隨時溝通

具體代碼實現能夠參考Github