LeetCode *** 240. 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

.

分析：

分治+二分搜索

代码：

class Solution {
public:
int myt;
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if(matrix.size()<1)return false;
myt=target;
return search(matrix,0,matrix.size()-1,0,matrix[0].size()-1);
}

bool search(vector<vector<int>>& matrix,int i1,int i2,int j1,int j2){
if(matrix[i1][j1]>myt)return false;
if(i1==i2&&j1==j2){
if(matrix[i1][j1]==myt)return true;
else return false;
}

int midi=(i1+i2)/2;
int midj=(j1+j2)/2;

if(matrix[midi][midj]<myt){
if(midi<i2&&midj<j2)return search(matrix,midi+1,i2,midj+1,j2)||search(matrix,i1,midi,midj+1,j2)||search(matrix,midi+1,i2,j1,midj);
else if(midi<i2)return search(matrix,midi+1,i2,j1,midj);
else if(midj<j2)return search(matrix,i1,midi,midj+1,j2);
}

else {
if(midi<i2&&midi>i1&&midj>j1&&midj<j2)return search(matrix,i1,midi,j1,midj)||search(matrix,midi+1,i2,j1,midj-1)||search(matrix,i1,midi-1,midj-1,j2);
else if(midi<i2&&midj>j1)return search(matrix,i1,midi,j1,midj)||search(matrix,midi+1,i2,j1,midj-1);
else if(midi>i1&&midj<j2)return search(matrix,i1,midi,j1,midj)||search(matrix,i1,midi-1,midj-1,j2);
else return search(matrix,i1,midi,j1,midj);
}

}
};