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:

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[
  [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.

思路

For example, we start searching the matrix from top right corner, initialize the current position to top right corner, if the target is greater than the value in current position, then the target can not be in entire row of current position because the row is sorted, if the target is less than the value in current position, then the target can not in the entire column because the column is sorted too. We can rule out one row or one column each time, so the time complexity is O(m+n).

In my code, I start from the bottom left, and it’s the same as the following reference.https://discuss.leetcode.com/topic/20064/my-concise-o-m-n-java-solution

We can also use binary search m or n times, the complexity will reach O(mlogn) or O(nlogm).

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