LeetCode -- Search a 2D Matrix II
题目描述：
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.
本题看起来是Search a 2D Matrix的扩展问题，其实和第一题的思路完全不一样。比较精简的思路是：
1.从左下开始找，如果matrix[r,c]（r为行，c为列） < target，c++；否则r--。
2.直到r或c越界为止。
实现代码：
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.
本题看起来是Search a 2D Matrix的扩展问题，其实和第一题的思路完全不一样。比较精简的思路是：
1.从左下开始找，如果matrix[r,c]（r为行，c为列） < target，c++；否则r--。
2.直到r或c越界为止。
实现代码：
public class Solution {
public bool SearchMatrix(int[,] matrix, int target)
{
var rowLen = matrix.GetLength(0);
var colLen = matrix.GetLength(1);
var row = rowLen - 1;
var col = 0;
while(col <= colLen - 1 && row >= 0){
if(matrix[row,col] == target){
return true;
}
else if(matrix[row,col] < target){
col ++;
}
else{
row --;
}
}
return false;
}
}