【LeetCode】240. Search a 2D Matrix II

Difficulty:medium

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Description

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.

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.

Intuition

refer to 二维数组中的查找 

Solution

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

　　

Complexity

Time complexity : O(m+n)

Space complexity : O(1)

 

What I've learned

1. Ought to have a good command of the thought in this problem.

 More:【目录】LeetCode Java实现