LintCode 38. Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of it.

This matrix has the following properties:

Integers in each row are sorted from left to right.

Integers in each column are sorted from up to bottom.

No duplicate integers in each row or column.

Have you met this question in a real interview?

Yes

Example

Consider the following matrix:

[
[1, 3, 5, 7],
[2, 4, 7, 8],
[3, 5, 9, 10]
]

Given target =

3

, return

2

.

Challenge

O(m+n) time and O(1) extra space

解题思路：

从左下角往右上角找，若是小于target就往右找，若是大于target就往上找。时间复杂度O(m+n) n 为行数，m为列数。

定义count 计数。

与 Leetcode 240. Search a 2D Matrix II 类似。

Java code:

public class Solution {
/**
* @param matrix: A list of lists of integers
* @param: A number you want to search in the matrix
* @return: An integer indicate the occurrence of target in the given matrix
*/
public int searchMatrix(int[][] matrix, int target) {
//check corner case
if (matrix == null || matrix.length == 0) {
return 0;
}
if (matrix[0] == null || matrix[0].length == 0) {
return 0;
}
//find from bottom left to top right
int n = matrix.length; //row
int m = matrix[0].length; //column
int x = n - 1;
int y = 0;
int count = 0;
while (x >= 0 && y < m) {
if (matrix[x][y] < target) {
y++;
} else if (matrix[x][y] > target) {
x--;
} else {
count++;
x--;
y++;
}
}
return count;
}
}

Reference:

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