[leetcode]74. Search a 2D Matrix(Java)

https://leetcode.com/problems/search-a-2d-matrix/#/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.

For example,

Consider the following matrix:

[
[1,   3,  5,  7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]

Given target = 

3

, return 

true

.

package go.jacob.day629;

public class Demo1 {
/*
* 最优解法：Binary search on an ordered matrix
* 二分查找
* Runtime: 0 ms.Your runtime beats 75.27 % of java submissions
*/
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
return false;
int m = matrix.length, n = matrix[0].length;
int start = 0, end = m * n - 1;
// 二分查找
while (start <= end) {
int mid = start + (end - start) / 2;
int value = matrix[mid / n][mid % n];
if (target > value) {
start = mid + 1;
} else if (target < value)
end = mid - 1;
else
return true;
}
return false;
}

/*
* 解法二： Solution by me Runtime: 1 ms.Your runtime beats 6.66 % of java
* submissions. 刚开始指向右上角元素，如果target大于该元素，下移；如果target小于该元素，左移；
*/
public boolean searchMatrix_1(int[][]
4000
matrix, int target) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
return false;
int m = 0, n = matrix[0].length - 1;
// 避免角标越界
while (m >= 0 && m < matrix.length && n >= 0 && n < matrix[0].length) {
if (target > matrix[m]
)
m++;
else if (target < matrix[m]
)
n--;
else
return true;
}
return false;

}
}