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How do I search for a number in a 2d array sorted left to right and top to bottom? (21 answers)

Closed 5 years ago.

页: 1

1   5   7    9
4   6   10   15
8   11  12   19
14  16  18   21

如何在<条码>O(x+y)中查找这一矩阵?

我被问及这个问题,但不能说出。想知道是否可以这样做。

Answer 1

Start at the last element of the first row(top-right corner).
Compare it with the key. We have 3 cases:

如果它们是平等的,我们就这样做了。
If key is greater than that element then it means key cannot be present in that row so move the search to the element below it.
If key is less than that element then it means key could be present in that row towards left and cannot be present in the column further down, so move the search to the element left of it.

坚持这样做,直到你找到这个要素,或者你不能再走(根本不存在)。

Pseudo Code:

Let R be number of rows
Let C be number of columns

Let i = 0
Let j = C-1

found = false
while( i>=0 && i<R) && (j>=0 && j<C) )
   if (matrix[i][j] == key )
      found = true
      break
   else if( matrix[i][j] > key )
       j--
   else if( matrix[i][j] < key )
       i++
end-while

Answer 2

在4个次级矩阵中提供矩阵。如果次矩阵的底层权利低于关键,则予以放弃。如果一个子矩阵的顶部大于钥匙,就予以抛弃。答复其余分矩阵的分选程序。

[Update] For some pseudo code (and a discussion of complexity) see Jeffrey L Whitledge s answer of this question.

Answer 3

// the matrix is like this, from left to right is ascending, and
// from down to up is ascending, but the second row start is not always bigger than the first row s end, which is diff from [leetcode]https://oj.leetcode.com/problems/search-a-2d-matrix/
// 1   5   7    9
// 4   6   10   15
// 8   11  12   19
// 14  16  18   21
// time complexity is O(x+y), x is the count of row, and y is the count of column

public boolean searchMatrix2(int[][] matrix, int target) {
    int rowCount = matrix.length;
    if(rowCount == 0) return false;

    int colCount = matrix[0].length;
    if(colCount == 0) return false;

    //first find the target row, needs O(x)
    int targetRow = 0;
    while(targetRow < rowCount-1 && matrix[targetRow+1][0] <= target) {
        targetRow++;
    }
    //than find the target in the target row, needs O(y), so the total is O(x)+O(y)
    boolean result = false;
    for(int i = 0; i < colCount; i ++) {
        if(matrix[targetRow][i] == target) {
            result = true;
            break;
        }
    }
    return result;
}

实际上,我们可以使用双轨搜索器,首先通过双轨搜索找到目标行,然后通过双轨搜索在浏览中找到目标,因此时间复杂性是O(lgx)+O(lgy),即O(lgx + lgy),即O(x+y)。

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