【leetcode】955. Delete Columns to Make Sorted II

题目如下：

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.

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"].

Suppose we chose a set of deletion indices D such that after deletions, the final array has its elements in lexicographic order (A[0] <= A[1] <= A[2] ... <= A[A.length - 1]).

Return the minimum possible value of D.length.

 

Example 1:

Input: ["ca","bb","ac"]
Output: 1
Explanation: 
After deleting the first column, A = ["a", "b", "c"].
Now A is in lexicographic order (ie. A[0] <= A[1] <= A[2]).
We require at least 1 deletion since initially A was not in lexicographic order, so the answer is 1.

Example 2:

Input: ["xc","yb","za"]
Output: 0
Explanation: 
A is already in lexicographic order, so we don't need to delete anything.
Note that the rows of A are not necessarily in lexicographic order:
ie. it is NOT necessarily true that (A[0][0] <= A[0][1] <= ...)

Example 3:

Input: ["zyx","wvu","tsr"]
Output: 3
Explanation: 
We have to delete every column.

Note:

1. 1 <= A.length <= 100
2. 1 <= A[i].length <= 100

解题思路：这题在【leetcode】944. Delete Columns to Make Sorted 的基础上提升了难度，【leetcode】944. Delete Columns to Make Sorted 只要求了A中所有字符串在同一索引的字符是升序的，而本题是要求字符串本身是升序的。我的解法是对A中所有字符串进行前缀比较，前缀的区间起始是0，重点是end （ 0 <= end < 字符串长度）。如果前缀比较中发现不满足升序条件，那么删除掉end位置的所有元素；否则end += 1，继续往后比较。

代码如下：

class Solution(object):
    def minDeletionSize(self, A):
        """
        :type A: List[str]
        :rtype: int
        """
        end = 0
        res = 0
        while end < len(A[0]):
            for i in range(len(A)-1):
                #print A[i][:end+1],A[i+1][:end+1]
                if A[i][:end+1] > A[i+1][:end+1]:
                    res += 1
                    #delete end + 1
                    for j in range(len(A)):
                        A[j] = A[j][:end] + A[j][end+1:]
                    end -= 1
                    break
                else:
                    continue
            end += 1
        return res