74. 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.

由于这道题里给出了限定条件，每一行的第一个元素一定比上一行最后一个元素大，所以整个二维数组可以被看成是一个被切分了几段的排序的一维数组。所以就可以用二分法搜索了。

class Solution {public:    bool searchMatrix(vector<vector<int> > &matrix, int target) {        if (matrix.empty() || matrix[0].empty()) return false;        if (target < matrix[0][0] || target > matrix.back().back()) return false;        int m = matrix.size(), n = matrix[0].size();        int left = 0, right = m * n - 1;        while (left <= right) {            int mid = (left + right) / 2;            if (matrix[mid / n][mid % n] == target) return true;            else if (matrix[mid / n][mid % n] < target) left = mid + 1;            else right = mid - 1;        }        return false;    }};