Leetcode240. Search a 2D Matrix II

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）.