leetcode 74|240. Search a 2D Matrix 1|II

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.

从右上到左下，依次检查当前元素cur，如果target>cur，那么cur所在的列都不用检查了，列向左推进一步。如果target<cur，那么cur所在的行都不用检查了，行向下推荐一步。

如果target==cur，返回cur。这个过程是个从右上走到左下的过程。最多走N+M步（N,M是边长），所以，时间复杂度为O(N+M)

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target)
{
//右上往左下找
if(matrix.empty())
return 0;
int row_num = matrix.size();
int col_num = matrix[0].size();

int row = 0;
int col = col_num-1;

while (row < row_num && col >= 0)
{
if (target > matrix[row][col])
row++;
else if (target < matrix[row][col])
col--;
else
return 1;
}
return 0;
}
};

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.

方法也是 右上到左下搜索！

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target)
{
if (matrix.empty())  return 0;
int row = 0;
int col = matrix[0].size() - 1;
while(row < matrix.size() && col >= 0)
{
if ( matrix[row][col] == target )
return 1;
else if (matrix[row][col] < target)
row++;
else
col--;
}
return 0;
}
};