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.

题意：先求出在哪行，问题等价于求第一个小于等于target的数.
然后用正常的二分查找。

class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        if(matrix.empty() || matrix[0].empty()) return false;        int mid, l, r, line = matrix.size();        l = 0, r = matrix.size() - 1;        while(l <= r){            mid = l + (r - l) / 2;            if(matrix[mid][0] == target){                return true;            } else if(matrix[mid][0] > target){                r = mid - 1;            } else if(matrix[mid][0] < target){                l = mid + 1;            }         }        if(r < 0) return false;        else line = r;        l = 0, r = matrix[0].size() - 1;        while(l <= r){            mid = l + (r - l) / 2;            if(matrix[line][mid] == target)                return true;            else if(matrix[line][mid] > target){                r = mid - 1;            } else if(matrix[line][mid] < target)                l = mid + 1;        }        return false;    }};