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 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.
这道题以前做过，从右上角开始，如果数字小于target，所在行不可能有target这个数，如果数字大于target，所在列不可能有target这个数，一开始写代码的时候会报错，错误为:Runtime Error Message： reference binding to null pointer of type ‘struct value_type’ 报错的时候的代码如下：
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { int m = matrix.size(); int n = matrix[0].size();//这里没有判断m是否存在 if(m == 0 || n == 0) return false; int i,j; bool is = false; i = 0;j = n-1; while(i < m && j >= 0) { if(matrix[i][j] == target) return true; else if(matrix[i][j] < target) i++; else if(matrix[i][j] > target) j--; } return is; } };一开始不知道什么意思，看了别人的代码才知道是判断m成立应该放在前面：
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { int m = matrix.size(); if(m == 0) return false; int n = matrix[0].size(); int i,j; bool is = false; i = 0;j = n-1; while(i < m && j >= 0) { if(matrix[i][j] == target) return true; else if(matrix[i][j] < target) i++; else if(matrix[i][j] > target) j--; } return is; } };这里代码的复杂度为 O（m+n）.
