[LeetCode] Search a 2D Matrix II

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

.

解题思路：

咋一看，还不知如何下手，因为这样不能像此前Search a 2D Matrix那样，将二维坐标转化成一维坐标。其实可以从二维的角度来将矩阵划分。



矩阵中心为9，其中左上角的所有元素均小于等于9，右下角的元素均大于9。因此，我们可以每次对比矩阵中心元素，然后排出左上角或右下角区域即可。下面是代码：

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();
if(n<=0){
return false;
}
return searchMatrixHelper(matrix, 0, m-1, 0, n-1, target);
}

bool searchMatrixHelper(vector<vector<int>>& matrix, int startM, int endM, int startN, int endN, int target){
if(startM > endM || startN > endN){
return false;
}
int middleM = (startM + endM) / 2;
int middleN = (startN + endN) / 2;
if(matrix[middleM][middleN]==target){
return true;
}else if(matrix[middleM][middleN]<target){
return searchMatrixHelper(matrix, startM, endM, middleN + 1, endN, target) ||
searchMatrixHelper(matrix, middleM + 1, endM, startN, middleN, target);
}else{
return searchMatrixHelper(matrix, startM, middleM - 1, startN, endN, target) ||
searchMatrixHelper(matrix, middleM, endM, startN, middleN - 1, target);
}
}
};