Search a 2D Matrix

Question

Problem Statement

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

Consider the following matrix:

[
    [1, 3, 5, 7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
]

Given target = 3, return true.

Challenge

O(log(n) + log(m)) time

题解 - 一次二分搜索 V.S. 两次二分搜索

  • 一次二分搜索 - 由于矩阵按升序排列,因此可将二维矩阵转换为一维问题。对原始的二分搜索进行适当改变即可(求行和列)。时间复杂度为 $$O(log(mn))=O(log(m)+log(n))$$
  • 两次二分搜索 - 先按行再按列进行搜索,即两次二分搜索。时间复杂度相同。

一次二分搜索

Python

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class Solution:
def search_matrix(self, matrix, target):
# Find the first position of target
if not matrix or not matrix[0]:
return False
m, n = len(matrix), len(matrix[0])
st, ed = 0, m * n - 1

while st + 1 < ed:
mid = (st + ed) / 2
if matrix[mid / n][mid % n] == target:
return True
elif matrix[mid / n][mid % n] < target:
st = mid
else:
ed = mid
return matrix[st / n][st % n] == target or \
matrix[ed / n][ed % n] == target

C++

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class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.empty() || matrix[0].empty()) return false;

int ROW = matrix.size(), COL = matrix[0].size();
int lb = -1, ub = ROW * COL;
while (lb + 1 < ub) {
int mid = lb + (ub - lb) / 2;
if (matrix[mid / COL][mid % COL] < target) {
lb = mid;
} else {
if (matrix[mid / COL][mid % COL] == target) return true;
ub = mid;
}
}
return false;
}
};

Java

lower bound 二分模板。

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public class Solution {
/**
* @param matrix, a list of lists of integers
* @param target, an integer
* @return a boolean, indicate whether matrix contains target
*/
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0 || matrix[0] == null) {
return false;
}

int ROW = matrix.length, COL = matrix[0].length;
int lb = -1, ub = ROW * COL;
while (lb + 1 < ub) {
int mid = lb + (ub - lb) / 2;
if (matrix[mid / COL][mid % COL] < target) {
lb = mid;
} else {
if (matrix[mid / COL][mid % COL] == target) {
return true;
}
ub = mid;
}
}

return false;
}
}

源码分析

仍然可以使用经典的二分搜索模板(lower bound),注意下标的赋值即可。

  1. 首先对输入做异常处理,不仅要考虑到matrix为null,还要考虑到matrix[0]的长度也为0。
  2. 由于 lb 的变化处一定小于 target, 故在 else 中判断。

复杂度分析

二分搜索,$$O(\log mn)$$.

两次二分法

Python

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class Solution:
def search_matrix(self, matrix, target):
if not matrix or not matrix[0]:
return False

# first pos >= target
st, ed = 0, len(matrix) - 1
while st + 1 < ed:
mid = (st + ed) / 2
if matrix[mid][-1] == target:
st = mid
elif matrix[mid][-1] < target:
st = mid
else:
ed = mid
if matrix[st][-1] >= target:
row = matrix[st]
elif matrix[ed][-1] >= target:
row = matrix[ed]
else:
return False

# binary search in row
st, ed = 0, len(row) - 1
while st + 1 < ed:
mid = (st + ed) / 2
if row[mid] == target:
return True
elif row[mid] < target:
st = mid
else:
ed = mid
return row[st] == target or row[ed] == target

源码分析

  1. 先找到first position的行, 这一行的最后一个元素大于等于target
  2. 再在这一行中找target

复杂度分析

二分搜索, $$O(\log m + \log n)$$




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