Search a 2D Matrix II

Question

Problem Statement

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.

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

题解 - 自右上而左下

  1. 复杂度要求——O(m+n) time and O(1) extra space,同时输入只满足自顶向下和自左向右的升序,行与行之间不再有递增关系,与上题有较大区别。时间复杂度为线性要求,因此可从元素排列特点出发,从一端走向另一端无论如何都需要m+n步,因此可分析对角线元素。
  2. 首先分析如果从左上角开始搜索,由于元素升序为自左向右和自上而下,因此如果target大于当前搜索元素时还有两个方向需要搜索,不太合适。
  3. 如果从右上角开始搜索,由于左边的元素一定不大于当前元素,而下面的元素一定不小于当前元素,因此每次比较时均可排除一列或者一行元素(大于当前元素则排除当前行,小于当前元素则排除当前列,由矩阵特点可知),可达到题目要求的复杂度。

在遇到之前没有遇到过的复杂题目时,可先使用简单的数据进行测试去帮助发现规律。

Python

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class Solution:
"""
@param matrix: An list of lists of integers
@param target: An integer you want to search in matrix
@return: An integer indicates the total occurrence of target in the given matrix
"""
def searchMatrix(self, matrix, target):
if not matrix or not matrix[0]:
return 0
occur = 0
row, col = 0, len(matrix[0])-1
while row < len(matrix) and col >= 0:
if matrix[row][col] == target:
occur += 1
col -= 1
elif matrix[row][col] < target:
row += 1
else:
col -= 1
return occur

C++

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class Solution {
public:
/**
* @param matrix: A list of lists of integers
* @param target: An integer you want to search in matrix
* @return: An integer indicate the total occurrence of target in the given matrix
*/
int searchMatrix(vector<vector<int> > &matrix, int target) {
if (matrix.empty() || matrix[0].empty()) {
return 0;
}

const int ROW = matrix.size();
const int COL = matrix[0].size();

int row = 0, col = COL - 1;
int occur = 0;
while (row < ROW && col >= 0) {
if (target == matrix[row][col]) {
++occur;
--col;
} else if (target < matrix[row][col]){
--col;
} else {
++row;
}
}

return occur;
}
};

Java

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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) {
int occurrence = 0;
if (matrix == null || matrix[0] == null) {
return occurrence;
}

int row = 0, col = matrix[0].length - 1;
while (row >= 0 && row < matrix.length && col >= 0 && col < matrix[0].length) {
if (matrix[row][col] == target) {
occurrence++;
col--;
} else if (matrix[row][col] > target) {
col--;
} else {
row++;
}
}

return occurrence;
}
}

源码分析

  1. 首先对输入做异常处理,不仅要考虑到matrix为空串,还要考虑到matrix[0]也为空串。
  2. 注意循环终止条件。
  3. 在找出target后应继续向左搜索其他可能相等的元素,下方比当前元素大,故排除此列。

严格来讲每次取二维矩阵元素前都应该进行 null 检测。

复杂度分析

由于每行每列遍历一次,故时间复杂度为 \[O(m + n)\].

Reference

Searching a 2D Sorted Matrix Part II | LeetCode




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