Leetcode 74. Search a 2D Matrix

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

.

解题思路：

做两次binary search, 先找在哪一行，即比较第一列，last number <= target.

找到那一行了，再用binary search找哪一列. 时间复杂度还是O(logn)

Java code:

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
//binary search twice, first use first column to find in which row
//then go to the row, find in which column
if(matrix == null || matrix.length == 0) {
return false;
}
if(matrix[0] == null || matrix[0].length == 0) {
return false;
}
int row = matrix.length;
int column = matrix[0].length;

//find the row index, the last number <= target
int start = 0;
int end = row - 1;
while(start + 1 < end) {
int mid = start + (end - start) / 2;
if(matrix[mid][0] == target) {
return true;
}else if(matrix[mid][0] < target) {
start = mid;
}else {
end = mid;
}
}
if (matrix[end][0] <= target) {
row = end;
} else if (matrix[start][0] <= target) {
row = start;
} else {
return false;
}

//find the column index, the number equal to target
start = 0;
end = column - 1;
while(start + 1 < end) {
int mid = start + (end - start) / 2;
if(matrix[row][mid] == target) {
return true;
}else if(matrix[row][mid] < target) {
start = mid;
}else {
end = mid;
}
}
if(matrix[row][start] == target){
return true;
}else if (matrix[row][end] == target) {
return true;
}else {
return false;
}
}
}

Reference:

1. http://www.jiuzhang.com/solutions/search-a-2d-matrix/