Search a 2D Matrix

题目：

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.

Example: Consider the following matrix:

[

[1, 3, 5, 7],

[10, 11, 16, 20],

[23, 30, 34, 50]

]

分析：

这道题目主要有意思的地方在于一个2D坐标到1D坐标的转换，如果说有一个m x n的矩阵，那矩阵的第k个元素，其实是矩阵中(k / n, k % n)位置上的那个元素。

解法：

class Solution {
public:
    /**
     * @param matrix, a list of lists of integers
     * @param target, an integer
     * @return a boolean, indicate whether matrix contains target
     */
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        // write your code here
        if (matrix.size() == 0) {
            return false;
        }
        if (matrix[0].size() == 0) {
            return false;
        }

        int start = 0;
        int m = matrix.size();
        int n = matrix[0].size();
        int end = m * n - 1;

        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid / n][mid % n] == target) {
                return true;
            } else if (matrix[mid / n][mid % n] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }

        if (matrix[start / n][start % n] == target) {
            return true;
        }

        if (matrix[end / n][end % n] == target) {
            return true;
        }

        return false;
    }
};