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.

 

 

Analyse: binary search. 
Runtime: 12ms.
 1 class Solution {
 2 public:
 3     bool searchMatrix(vector<vector<int>>& matrix, int target) {
 4         if(matrix.empty() || matrix[0].empty()) return false;
 5         
 6         int m = matrix.size(), n = matrix[0].size();
 7         // target the row, low would be the target
 8         int low = 0, high = m - 1;
 9         while(low < high) {
10             int mid = low + (high - low) / 2;
11             if(matrix[mid][n - 1] > target) high = mid;
12             else if(matrix[mid][n - 1] < target) low = mid + 1;
13             else return true;
14         }
15         
16         // target the column
17         int start = 0, end = n - 1;
18         while(start < end) {
19             int mid = start + (end - start) / 2;
20             if(matrix[low][mid] > target) end = mid;
21             else if(matrix[low][mid] < target) start = mid + 1;
22             else return true;
23         }
24         return matrix[low][start] == target;
25     }
26 };

 

posted @ 2016-08-11 01:45  amazingzoe  阅读(123)  评论(0)  编辑  收藏  举报