leetcode -- 74. Search a 2D Matrix

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题目描述

题目难度：Medium
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.

• Example 1:

Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true

• Example 2:

Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false

AC代码

这个题目不是很难，但是有一个巧妙的解法，根据题目所给数组的特性，可以把二维数组当成有序的一维数组，然后在其上的查找就可以用binary-search。

class Solution {public boolean searchMatrix(int[][] matrix, int target) {if(matrix == null || matrix.length == 0) return false;int row = matrix.length, col = matrix[0].length;int low = 0, high = row * col - 1;while(low <= high){int mid = low + ((high - low) >> 1);//((high - low) >> 1)最外层要加括号，由于优先级的问题，3 + 8 >> 1 其实等于 5.int midVal = matrix[mid / col][mid % col]; if( midVal == target) return true;else if(midVal > target) high = mid - 1;else low = mid + 1;}return false;}
}

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