Search a 2D Matrix

最新推荐文章于 2022-08-05 12:18:43 发布

原创最新推荐文章于 2022-08-05 12:18:43 发布 · 846 阅读

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本文介绍了一种高效的算法，用于在一个m x n的二维矩阵中查找特定值。该矩阵每一行的元素从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章提供了详细的算法实现步骤及示例。

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.

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
        int rowIndex = getRowIndex(matrix, target);
        return exists(matrix, rowIndex, target);
        
    }
    
    public int getRowIndex(int[][] matrix, int target) {
        int i = 0;
        int j = matrix.length - 1;
        
        while (i <= j) {
            int mid = i + (j - i) / 2;
            if (matrix[mid][0] == target) {
                return mid;
            } else if (matrix[mid][0] > target) {
                j--;
            } else {
                i++;
            }
        }
        return j;
    }
    
    public boolean exists(int[][] matrix, int rowIndex, int target) {
        if (rowIndex < 0 || rowIndex >= matrix.length) return false;
        int i = 0;
        int j = matrix[0].length - 1;
        while (i <= j) {
            int mid = i + (j - i) / 2;
            if (matrix[rowIndex][mid] == target) {
                return true;
            } else if (matrix[rowIndex][mid] > target) {
                j--;
            } else {
                i++;
            }
        }
        return false;
    }
}