240. Search a 2D Matrix II

/*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 in ascending from left to right.Integers in each column are sorted in ascending from top to bottom.For example,Consider the following matrix:[  [1,   4,  7, 11, 15],  [2,   5,  8, 12, 19],  [3,   6,  9, 16, 22],  [10, 13, 14, 17, 24],  [18, 21, 23, 26, 30]]Given target = 5, return true.Given target = 20, return false.由于每一行从左到右、每一列从上到下都是升序，我们可以从矩阵的第一行最右侧开始查找，如果matrix[row][column]==target则返回true;如果matrix[row][column]<target,则说明这一行的所有数都小于target，则直接转到下一行；如果matrix[row][column]>target,则说明这一列所有的数都大于target，则向左移动一列。*/class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        if(matrix.size()==0 || matrix[0].size()==0) return false;        int row=0,column=matrix[0].size()-1;        while(row<matrix.size() && column>=0)        {            if(target == matrix[row][column]) return true;            else if(target > matrix[row][column]) row++;            else column--;        }        return false;    }};