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搜索算法，用于在一特殊类型的二维数组中查找特定数值。该数组每一行从左到右递增排序，且每行首元素大于前一行末元素。通过先定位目标值所在行再使用二分查找实现快速检索。

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:

题意：在一个二维数组（其一维形式属于一个递增的序列）需找一个数，存在则返回TRUE，否则返回FALSE。

思路：先找到该数位于素组的哪一行，然后在这行中用二分查找。

代码：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length,n = matrix[0].length, index = -1;
		for(int i = 0; i < m; i++)
		{
			if(target >= matrix[i][0] && target <= matrix[i][n - 1])
			{
				index = i;
				break;
			}
		}
		if(index == -1)
			return false;
        int left = 0, right = n - 1;
        while(left <= right)
        {
        	int mid = (left + right) /2;
        	if(matrix[index][mid] == target)
        		return true;
        	else if(matrix[index][mid] > target)
        		right = mid - 1;
        	else
        		left = mid + 1;
        }
        return false;
    }
}