LeetCode 74. Search a 2D Matrix（搜索二维矩阵）

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.

---

题目标签：Array

　　题目给了我们一个有序的矩阵，让我们找target是否在矩阵中。首先我们分析，如果我们知道了target 在哪一行的话，直接用binary search 就可以了。所以第一步可以用binary search 来搜索第一列，确认target在哪一行。在用一次二分法对于那一行，找出target。

Java Solution:

Runtime beats 71.85%

完成日期：07/23/2017

关键词：Array

关键点：用二分法两次

 public class Solution
 {
     public boolean searchMatrix(int[][] matrix, int target)
     {
         if(matrix.length == 0 || matrix[0].length == 0)
             return false;
         if(target < matrix[0][0] || target > matrix[matrix.length-1][matrix[0].length-1])
             return false;
         // first determine which row the target is in
         int top = 0;
         int bottom = matrix.length-1;

         while(top <= bottom)
         {
             int mid = top + (bottom - top) / 2;

             if(matrix[mid][0] == target)
                 return true;
             else if(matrix[mid][0] > target)
                 bottom = mid - 1;
             else
                 top = mid + 1;
         }

         int rowNum = bottom;

         // use binary search for this row
         int left = 0;
         int right = matrix[0].length-1;

         while(left <= right)
         {
             int mid = left + (right - left) / 2;

             if(matrix[rowNum][mid] == target)
                 return true;
             else if(matrix[rowNum][mid] < target)
                 left = mid + 1;
             else
                 right = mid - 1;

         }

         return false;
     }
 }

参考资料：

http://www.cnblogs.com/grandyang/p/4323301.html

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