LeetCode: Search a 2D Matrix 解题报告

Search a 2D Matrix

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.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

SOLUTION 1:

采用经典的二分法模板

while (left <= right) {
            int mid = left + (right - left) / 2;
            
         
            int n = matrix[row][col];
            
            if (n == target) {
                ...
            } else if (n < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        
        int rows = matrix.length;
        int cols = matrix[0].length;
        
        int num = rows * cols;
        
        int left = 0;
        int right = num - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            int row = mid / cols;
            int col = mid % cols;
            
            int n = matrix[row][col];
            
            if (n == target) {
                return true;
            } else if (n < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return false;        
    }
}

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/divide2/SearchMatrix.java