lintcode-easy-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.

Consider the following matrix:

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

Given target = 3, return true.

和普通二分搜索是一样的

public class Solution {
    /**
     * @param matrix, a list of lists of integers
     * @param target, an integer
     * @return a boolean, indicate whether matrix contains target
     */
    public boolean searchMatrix(int[][] matrix, int target) {
        // write your code here

        if(matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0)
            return false;

        int m = matrix.length;
        int n = matrix[0].length;

        int size = m * n;

        int left = 0;
        int right = size - 1;

        while(left <= right){
            int mid = left + (right - left) / 2;

            int i = mid / n;
            int j = mid % n;

            if(matrix[i][j] == target)
                return true;
            else if(matrix[i][j] < target)
                left = mid + 1;
            else
                right = mid - 1;
        }

        return false;
    }
}