【LeetCode】74. Search a 2D Matrix

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 from left to right.
• The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

Intuition

regard the matrix as an array, and then use binary search.

　　matrix[x][y]=array[x*cols+y]

　　array[m]=matrix[m/cols][m%cols]

Solution

    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix==null || matrix.length<=0 || matrix[0].length<=0)
            return false;
        int rows=matrix.length;  //行数
        int cols=matrix[0].length;  //列数
        int low=0;
        int high=rows*cols-1;
        while(low<=high){
            int mid=(low+high)/2;
            if(matrix[mid/cols][mid%cols]==target){
                return true;
            }else if(matrix[mid/cols][mid%cols]>target){
                high=mid-1;
            }else if(matrix[mid/cols][mid%cols]<target){
                low=mid+1;
            }
        }
        return false;
    }

　　

Complexity

Time complexity : O(log(n*m))

Space complexity : O(1)

What I've learned

1. Ought to have a good command of the change between array <=> matrix, especially the change of their indexes

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