LeetCode 1428. Leftmost Column with at Least a One

原题链接在这里：https://leetcode.com/problems/leftmost-column-with-at-least-a-one/

题目：

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 

Constraints:

• rows == mat.length
• cols == mat[i].length
• 1 <= rows, cols <= 100
• mat[i][j] is either 0 or 1.
• mat[i] is sorted in non-decreasing order.

题解：

From the upper right corner, get the element.

If it is 1, then there could be 1s on its left, col--.

If it is 0, then there could be 1s on its down, row++. Why can't it move right, because we moved from the right to left, we already mark that position.

Time Complexity: O(m + n). m is the row count of matrix. n is column count of matrix.

AC Java:

/**
 * // This is the BinaryMatrix's API interface.
 * // You should not implement it, or speculate about its implementation
 * interface BinaryMatrix {
 *     public int get(int row, int col) {}
 *     public List<Integer> dimensions {}
 * };
 */

class Solution {
    public int leftMostColumnWithOne(BinaryMatrix binaryMatrix) {
        List<Integer> dimen = binaryMatrix.dimensions();
        int m = dimen.get(0);
        int n = dimen.get(1);
        int i = 0;
        int j = n - 1;
        int res = -1;
        while(i < m && j >= 0){
            if(binaryMatrix.get(i, j) == 0){
                i++;
            }else{
                res = j;
                j--;
            }
        }
        
        return res;
    }
}