Search a 2D Matrix 在有序二维矩阵查找数 @LeetCode

经典面试题，从右上角开始查找O(m+n)，可以不用二分法

package Level3;

/**
 * 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.
 *
 */
public class S74 {

	public static void main(String[] args) {

	}
	
	public boolean searchMatrix(int[][] matrix, int target) {
		
		// 取右上角的进行最初比较
		int x = 0, y = matrix[0].length-1;
		
		// 限定范围
		while(x>=0 && x=0 && y matrix[x][y]){
				x++;
			}else if(target < matrix[x][y]){		// 不必在这一列中查找
				y--;
			}
		}
		return false;
    }

}

Second try:

发现二分法的方法：

http://n00tc0d3r.blogspot.com/2013/05/search-2d-matrix.html

Another way to solve the problem is to think of the 2D matrix as a sorted 1D array and then apply the regular binary search on the 1D array. After computing the mid value, we calculate the corresponding row and column numbers based on mid.

public static boolean searchMatrix2(int[][] matrix, int target) {  
		   // treat the 2D matrix as a sorted 1D array and use binary search to find the target  
		   int low = 0, high = matrix.length * matrix[0].length - 1;  
		   while (low <= high) {  
		     int mid = (low + high) / 2;  
		     int row = mid / matrix[0].length, col = mid % matrix[0].length;  
		     if (target == matrix[row][col]) return true;  
		     else if (target < matrix[row][col]) high = mid - 1;  
		     else low = mid + 1;  
		   }  
		   
		   return false;  
	} 

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int rows = matrix.length, cols=matrix[0].length;
        if(matrix.length == 0){
            return false;
        }
        int i=0, j=matrix[0].length-1;
        
        while(i>=0 && i=0 && j target){
                j--;
            }else{
                i++;
            }
        }
        
        return false;
    }
}