【Leetcode】之Search a 2D Matrix

一.问题描述

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 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

.

二.我的解题思路

这道题还是比较简单的，就是二分法的变形。测试通过程序如下：

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int row_num = matrix.size();
if(row_num==0) return 0;
int line_num = matrix[0].size();
int total_num = row_num*line_num;
int st=0;int end=total_num-1;int mid;
while(st<end){
mid = (st+end)/2;
int curr_val = gen_val(matrix,mid);
if(curr_val==target) return 1;
if(curr_val<target) st=mid+1;
if(curr_val>target) end=mid-1;
}
if(st==end){
int curr_val = gen_val(matrix,st);
if(curr_val==target) return 1;
else return 0;
}
return 0;

}

int gen_val(vector<vector<int>>& matrix, int idx){
int row_num = matrix.size();
int line_num = matrix[0].size();
int i,j;
i=idx/line_num;
j=idx-i*line_num;
return matrix[i][j];

}
};