資訊專欄INFORMATION COLUMN
摘要：題目就是給一個數(shù)組，把每一項的對應的組合成一個新的數(shù)組，再算出那些不是遞增的個數(shù)。例子我的算法其他算法思路跟我的差不多，只不過用了的
題目
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"], and the remaining columns of A are ["b","v"], ["e","y"], and ["f","z"]. (Formally, the c-th column is A[0, A1, ..., AA.length-1].)
Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order.
就是給一個數(shù)組， 把每一項的對應的index組合成一個新的數(shù)組，再算出那些不是遞增的個數(shù)。
Return the minimum possible value of D.length.
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.
Input: ["a","b"]
Output: 0
Explanation: D = {}
Input: ["zyx","wvu","tsr"]
Output: 3
Explanation: D = {0, 1, 2}
我的算法
var minDeletionSize = function(A) {
const b = A.map(v => v.split(""))
let len = A.length
let len2 = b[0].length
let n = 0
for (var i=0;i b[j+1][i]){
n++
break
}
}
}
return n
};
Runtime: 88 ms, faster than 64.27% of JavaScript online submissions for Delete Columns to Make Sorted.其他算法
Memory Usage: 43.9 MB, less than 7.69% of JavaScript online submissions for Delete Columns to Make Sorted.
class Solution:
def minDeletionSize(self, A):
return sum(any(a[j] > b[j] for a, b in zip(A, A[1:])) for j in range(len(A[0])))
思路跟我的差不多，只不過用了python的zip api
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