74. 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.

在一个m*n二维数组中，每一行从左到右递增，每一行的第一个元素比上一行最后一个元素大。

判断某个元素是否在 该数组中。

 

 1     public boolean searchMatrix(int[][] matrix, int target) {
 2         if (matrix.length == 0 || matrix[0].length == 0) return false;
 3         if (target < matrix[0][0] || target > matrix[matrix.length - 1][matrix[0].length - 1]) return false;
 4         int m = matrix.length;
 5         int n = matrix[0].length;
 6         int left = 0, right = m * n - 1;
 7         while (left < right) {
 8             int mid = left + (right - left) / 2;
 9             if (matrix[mid / n][mid % n] == target) return true;
10             else if (matrix[mid / n][mid % n] > target) right = mid - 1;
11             else left = mid + 1;
12         }
13         return matrix[left / n][left % n] == target;
14     }

 

posted @ 2017-10-16 10:59  daniel456  阅读(119)  评论(0)  编辑  收藏  举报