# 74.Search-a-2D-Matrix

## 题目地址

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

http://www.lintcode.com/problem/search-a-2d-matrix/description

## 题目描述

``````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``````

## 代码

Approach #1 Brute Force

``````class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] == target) {
return true;
}
}
}

return false;
}
}``````

Approach #2 Binary Search twice

``````public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
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, 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;
}

}

}``````

Approach 2: Binary Search Once

``````class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
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;

int start = 0, end = row * column - 1;
while (start <= end) {         // [[1]] 改成 start + 1 < end 结果错误
int mid = start + (end - start) / 2;
int number = matrix[mid / column][mid % column];
if (number == target) {
return true;
} else if (number > target) {
end = mid - 1;
} else {
start = mid + 1;
}
}

return false;
}
}``````

Last updated