LeetCode-Delete Columns to Make Sorted

最新推荐文章于 2019-04-07 09:09:48 发布

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最新推荐文章于 2019-04-07 09:09:48 发布

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分类专栏： LeetCode 文章标签： LeetCode Java

本文链接：https://blog.youkuaiyun.com/qq_24133491/article/details/84350515

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本文探讨了如何通过选择最小数量的删除索引，使字符串数组中的每列字符按非递减顺序排列的问题。提供了详细的例子和解决方案，包括Java代码实现。

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Description:
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

题意：给定一个字符串数组A和一个下标集合D，删除数组A中每个字符串在D中所有元素所对应下标位置处的字符；计算，最小长度的集合D，使得删除后的字符串数组A中所有字符串对应下标位置处的字符为增序排序；
例如：

for A = {“cba”, “daf”, “ghi”}
if D = {1}
then A = {“ca”, “df”, “gi”}
则字符串A中所有字符串对应下标位置处的字符为
- index₀: {‘c’, ‘d’, ‘g’}
- index₂: {‘a’, ‘f’, ‘i’}

满足所有index_i处的字符为增序排序，且此时集合D的长度最小（因为当集合长度为0时，index₁处的字符不满足增序排序）

解法：这道题的目的是想让我们求出最小长度的集合D，其实，我们只需要求出在字符串数组A中有多少个index_i处的字符不满足增序排序，其值就是所要求解的集合D的长度;

Java

class Solution {
    public int minDeletionSize(String[] A) {
        int cnt = 0;
        int strLen = A[0].length();
        for (int i = 0; i < strLen; i++) {
            boolean des = true;
            for (int j = 1; j < A.length; j++) {
                if (A[j].charAt(i) < A[j-1].charAt(i)) {
                    des = false;
                    break;
                }
            }
            cnt = des ? cnt : cnt + 1;
        }
        return cnt;
    }
}