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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