[leetcode]240. Search a 2D Matrix II

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

代码

1.

/*
	 * Author:郭方舟
	 * Problem：240—Search_a_2D_Matrix II
	 * Algorithm:遍历行，每行上的元素用二分查找 
	 * time：2017/03/16
	 * time complexity is O(mlog2(n))
	 * space complexity is O(1)
	 */
	  /*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.

	/*
	 * Author:郭方舟
	 * Problem：240—Search_a_2D_Matrix II
	 * Algorithm:刚开始把右上角的元素跟 target进行比较，如果它比target大，列数就减一 ；如果它比target小 ,行数就加一
	 * time：2017/03/16
	 * time complexity is O(m+n)
	 * space complexity is O(1)
	 * beats:48.57%
	 */
	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
		int i = 0 , j = n -1;
		while( i < m && j >=0){
			if(matrix[i][j] < target){
				i++;
			}else if(matrix[i][j] > target){
				j--;
			}else{
				return true;
			}
		}
		return false;
	}	
}