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 · 519 阅读

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本文介绍了一种高效的算法，用于在一个特殊格式的二维矩阵中查找特定数值。该矩阵每一行的整数从左到右递增排序，并且每行的第一个整数大于前一行的最后一个整数。文章提供了一个Java实现示例。

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) {
		// Start typing your Java solution below
		// DO NOT write main() function
		int row = matrix.length;
		if(row == 0)
			return false;
		int col = matrix[0].length;
		if(col == 0)
			return false;
		int i = 0, j = col - 1;
		while(i < row && j >= 0){
			if(matrix[i][j] == target)
				return true;
			else if(matrix[i][j] > target)
				j--;
			else
				i++;
		}
		return false;
	}
}

注意学习下杨氏矩阵，不同于此题