Leetcode: Search a 2D Matrix

    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.

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

class Solution
{
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target)
    {
        int rows = int(matrix.size());
        int row = 0;
        bool flag = false;//标志是否有某一行的第一个数字比给定的数字大
        for (int i = 0; i < rows; ++i)
        {
            if (matrix[i][0] == target) return true;
            if (matrix[i][0] > target)
            {
                row = i;
                flag = true;
                break;
            }
        }
        //这里是对特殊情况的处理：如果每一行的第一个数字都比给定数字小，那么说明给定数字有可能在最后一行
        //如果falg=true而且row=1，说明第一行的第一个数字比给定数字大，那么矩阵中肯定不存在给定数字，直接返回false
        if (flag && row == 0) return false;
        else if (!flag) row = rows;
        //下面是对矩阵第row - 1行进行二分查找的过程
        int left = 0;
        int right = int(matrix[row - 1].size()) - 1;
        int middle = right / 2;
        while (left <= right)
        {
            middle = (left + right) / 2;
            if (matrix[row - 1][middle] > target) right = middle - 1;
            else if (matrix[row - 1][middle] < target) left = middle + 1;
            else return true;
        }
        return false;
    }
};