(This problem is an interactive problem.)
A row-sorted binary matrix means that all elements are 0
or 1
and each row of the matrix is sorted in non-decreasing order.
Given a row-sorted binary matrix binaryMatrix
, return the index (0-indexed) of the leftmost column with a 1 in it. If such an index does not exist, return -1
.
You can‘t access the Binary Matrix directly. You may only access the matrix using a BinaryMatrix
interface:
BinaryMatrix.get(row, col)
returns the element of the matrix at index (row, col)
(0-indexed).BinaryMatrix.dimensions()
returns the dimensions of the matrix as a list of 2 elements [rows, cols]
, which means the matrix is rows x cols
.Submissions making more than 1000
calls to BinaryMatrix.get
will be judged Wrong Answer. Also, any solutions that attempt to circumvent the judge will result in disqualification.
For custom testing purposes, the input will be the entire binary matrix mat
. You will not have access to the binary matrix directly.
Example 1:
Input: mat = [[0,0],[1,1]] Output: 0
Example 2:
Input: mat = [[0,0],[0,1]] Output: 1
Example 3:
Input: mat = [[0,0],[0,0]] Output: -1
Example 4:
Input: mat = [[0,0,0,1],[0,0,1,1],[0,1,1,1]] Output: 1
分析:
因为我们只知道每个row是sorted,但是colo并没有。最直接的办法是对于每个row, 返回是1的最小column值, 然后每个row比较一下就知道了。
1 class Solution { 2 public int leftMostColumnWithOne(BinaryMatrix binaryMatrix) { 3 List<Integer> dim = binaryMatrix.dimensions(); 4 int r = dim.get(0), c = dim.get(1); 5 int leftMost = c; 6 7 for (int i = 0; i < r; i++) { 8 int left = 0; 9 int right = c - 1; 10 while (left <= right) { 11 int mid = left + (right - left)/2; 12 if (binaryMatrix.get(i, mid) < 1) { 13 left = mid+1; 14 } 15 else { 16 right = mid - 1; 17 } 18 } 19 leftMost = leftMost > left? left : leftMost; 20 } 21 22 return leftMost == c ? -1 : leftMost; 23 } 24 }
1428. Leftmost Column with at Least a One
原文:https://www.cnblogs.com/beiyeqingteng/p/14493572.html