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Description:
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
题意:给定一个二维数组matrix和一个目标元素target,要求判断在数组matrix中是否存在目标元素target;并且,这里的二维数组按照行优先的话,是按照升序排序的(为利用二分查找提供了条件);
解法:对于一维数组来说,如果我们要找到一个目标元素,很容易的想到可以利用二分查找来实现;对于二维数组来说,其实它也可以看作一个一维数组,对于行数和列数分别为rows和cols的二维数组来说,与一维数组的转换为(假设此时要计算一维数组表示中pos处的元素在二维数组中的位置)
matrix[pos / cols][pos % cols]
因此,我们只需要像在一维数组中利用二分查找一样在二维数组中使用即可;
Java
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 low = 0;
int high = rows * cols - 1;
while (low <= high) {
int mid = (low + high) / 2;
if (matrix[mid / cols][mid % cols] == target) {
return true;
} else if (matrix[mid / cols][mid % cols] > target) {
high = mid - 1;
} else {
low = mid + 1;
}
}
return false;
}
}