【leetcode】Array——Search a 2D Matrix（74）

题目：

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

.

思路：两次binary search，第一次查询第一列，判断target可能出现在哪一行，然后再binary search该行。

第一次提交没过，因为没有考虑到如果target<matrix[0][0]，第一次binary search后确定的行会是-1，导致后面ArrayIndexOutOfBoundsException:
-1

代码：

public boolean searchMatrix(int[][] matrix, int target) {
int rows=matrix.length;
if(rows==0)
return false;
int lines=matrix[0].length;

//search first element of each row
int up=0,down=rows-1;
while(up<=down){
int mid = up + ((down-up)>>1);
if(matrix[mid][0]==target)
return true;
if(target<matrix[mid][0])
down=mid-1;
else
up=mid+1;
}
if(down==-1)
return false;
//search the row
int left=0,right=lines-1;
while(left<=right){
int mid = left+((right-left)>>1);
if(matrix[down][mid]==target)
return true;
if(target<matrix[down][mid])
right=mid-1;
else
left=mid+1;
}
return false;
}

改进：不要把matrix看成是二维数组

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0) {
return false;
}
int start = 0, rows = matrix.length, cols = matrix[0].length;
int end = rows * cols - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (matrix[mid / cols][mid % cols] == target) {
return true;
}
if (matrix[mid / cols][mid % cols] < target) {
start = mid + 1;
} else {
end = mid - 1;
}
}
return false;
}
}

leetcode链接：https://leetcode.com/discuss/10735/dont-treat-it-as-a-2d-matrix-just-treat-it-as-a-sorted-list