LeetCode Search a 2D Matrix(二分查找法)


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

.

题意：给出一个矩阵，第一行的元素从小到大排序，每一行的第一个元素比前一行的最后一个元素大，用有效的算法求是否可以找到数target

思路：以每一行来二分查找

代码如下

public class Solution {
private int lower_bound(int[] arrays, int target)
{
int left = 0, right = arrays.length;
int mid;

while (left < right)
{
mid = (left + right) >> 1;
if (arrays[mid] >= target) right = mid;
else
{
left = mid + 1;
}
}

return left;
}

public boolean searchMatrix(int[][] matrix, int target)
{
int m = matrix.length;
int n = m > 0 ? matrix[0].length : 0;

for (int i = 0; i < m; i++)
{
int left = lower_bound(matrix[i], target);
if (left >= n) continue;
if (matrix[i][left] == target) return true;
else return false;
}

return false;
}
}