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:

[
  [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.

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

反而更简单了，从右上角还是比较，target大就说明在下一row，target小就说明在前一column.同理也可以先比较左下角.

为什么右上角比较特殊呢？因为比他小的数都在他左边，比他大的数都在他下面。

同理左下角，比他小的都在上面，比他大的都在他右面。