leetcode笔记：240. search a 2d matrix ii

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

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[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

从数组的右上角开始，如果matrix[row][col] > target，则col--；若matrix[row][col] < target，则row++。因为，matrix[row][col] > target，则[col, n)的列都比target大；matrix[row][col] < target，则[0, row]的行都比target小。

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class  {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0 || 
           matrix[0] == null || matrix[0].length == 0) return false;
        
        int m = matrix.length, n = matrix[0].length;
        int row = 0, col = n - 1;
        while(row < m && col >= 0) {
            if(matrix[row][col] == target) {
                return true;
            } else if(matrix[row][col] > target) {
                col--;
            } else {
                row++;
            }
        }
        return false;
    }
}

因为每次扫描不是向左就是向下，因此时间复杂度为$$O(m+n)$$。