[Likou] 74. Search two-dimensional matrix <dichotomy>

[Likou] 74. Search two-dimensional matrix

Given an mxn matrix of integers satisfying the following two properties:

• The integers in each row are in non-decreasing order from left to right.
• The first integer in each row is greater than the last integer in the previous row.

Given an integer target, if target is in the matrix, return true; otherwise, return false.

Example 1:

1	3	5	7
10	11	16	20
23	30	34	60

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

1	3	5	7
10	11	16	20
23	30	34	60

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

提示：
m == matrix.length
n == matrix[i].length
1 <= m, n <= 100
-$10^4$ <= matrix[i][j], target <= $10^4$

answer

Dichotomy improvement, mapping a two-dimensional array to a one-dimensional array for dichotomy

public class Solution {
    
    
    public boolean searchMatrix(int[][] matrix, int target) {
    
    
        if (matrix == null || matrix.length == 0) {
    
    
            return false;
        }
        int row = matrix.length;
        int col = matrix[0].length;

        int left = 0;
        int right = row * col - 1;

        while (left <= right) {
    
    
            int mid = left + (right - left) / 2;

            // (x,y) --> x*col+y
            //反过来：一维转二维：matrix[mid/col][mid%col]
            int element = matrix[mid / col][mid % col];
            if (element == target) {
    
    
                return true;
            }
            else if (element > target) {
    
    
                right = mid - 1;
            }
            else {
    
    
                left = mid + 1;
            }
        }
        return false;
    }
}