[leetcode]74. Search a 2D Matrix

题目

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.

代码

1.

/*
 * Author:郭方舟
 * Problem：74—Search_a_2D_Matrix
 * Algorithm:二分查找
 * time：2017/03/15
 */
public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
      if(matrix ==  null || matrix.length == 0 || matrix[0].length == 0){
			return false;
		}
		int m = matrix.length;	  //行数
		int n = matrix[0].length; //列数
		
		for(int i = 0 ; i < m;i++){
			int start = 0,end = n-1;
			while(start <= end ){
				int mid = start + (end - start)/2;
				if(matrix[i][mid] == target){
					return true;
				}else if(matrix[i][mid] < target){
					start = mid +1;
				}else{
					end = mid -1;
				}
			}
		}
		return false;
	}
}

2.

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int n = matrix.size();
        int m = matrix[0].size();
        int l = 0, r = m * n - 1;
        while (l != r){
            int mid = (l + r - 1) >> 1;
            if (matrix[mid / m][mid % m] < target)
                l = mid + 1;
            else 
                r = mid;
        }
        return matrix[r / m][r % m] == target;
    }
};