Java for LeetCode 074 Search a 2D Matrix

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 from left to right.
• The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

解题思路：

二分查找即可，JAVA实现如下：

    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix.length == 0 || matrix[0].length == 0)
			return false;
		int up = 0, down = matrix.length-1, left = 0, right = matrix[0].length-1;
		while (up < down) {
			if(target>matrix[(up+down)/2][0]){
				up=(up+down)/2;
				if(target>matrix[up][matrix[0].length-1])
					up++;
				else break;
			}
			else if(target<matrix[(up+down)/2][0])
				down=(up+down)/2-1;
			else return true;
		}
		while(left<right){
			if(target>matrix[up][(left+right)/2])
				left=(left+right)/2+1;
			else if(target<matrix[up][(left+right)/2])
				right=(left+right)/2-1;
			else return true;
		}
		return target==matrix[up][left];
    }

当然，也可以只用两个指针表示，JAVA实现如下：

	public boolean searchMatrix(int[][] matrix, int target) {
		if (matrix.length == 0 || matrix[0].length == 0)
			return false;
        int l = 0, r = matrix.length * matrix[0].length - 1, mid = 0;
        while (l < r) {
            mid = l + (r - l) / 2;
            if (matrix[mid/matrix[0].length][mid%matrix[0].length] < target)
                l = mid + 1;
            else if (matrix[mid/matrix[0].length][mid%matrix[0].length] > target)
                r = mid;
            else return true;
        }
        return matrix[l/matrix[0].length][l%matrix[0].length] == target;
    }