109 lines
2.1 KiB
Markdown
109 lines
2.1 KiB
Markdown
# Leetcode Search-a-2D-Matrix
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#### 2022-06-13 15:33
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---
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##### Algorithms:
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#algorithm #binary_search
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##### Data structures:
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#DS #vector #vector_2d
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##### Difficulty:
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#leetcode #coding_problem #difficulty_easy
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/search-a-2d-matrix/submissions/)
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---
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### Problem
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Write an efficient algorithm that searches for a value `target` in an `m x n` integer matrix `matrix`. This matrix has the following properties:
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- Integers in each row are sorted from left to right.
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- The first integer of each row is greater than the last integer of the previous row.
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#### Examples
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**Example 1:**
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```markdown
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**Input:** matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
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**Output:** true
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```
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**Example 2:**
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```matrix
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**Input:** matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
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**Output:** false
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```
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#### Constraints
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- m == matrix.length
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- n == matrix[i].length
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- 1 <= m, n <= 100
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- -104 <= matrix[i][j], target <= 104
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### Thoughts
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Binary search algorithm, with simple 2-d to 1-d conversion
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> [!tip] Binary search while loops
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> In binary search, remember to check for:
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>
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> - l <= r
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> And use:
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> - l = mid + 1
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> - r = mid - 1
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> Init r as m \* n - 1 to avoid mid being OOB.
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### Solution
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==note how I calculated the mid, i, j, and how I changed r and l==
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```cpp
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class Solution {
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public:
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bool searchMatrix(vector<vector<int>> &matrix, int target) {
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int m = matrix.size();
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int n = matrix[0].size();
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int l = 0;
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// set r to len - 1 to avoid mid being bigger than len.
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int r = m * n - 1;
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int i; // the key to speed is to precaculate i and j, to save time.
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int j;
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int mid;
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do {
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mid = l + (r - l) / 2;
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i = mid / n;
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j = mid % n;
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if (target == matrix[i][j]) {
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return true;
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} else if (target < matrix[i][j]) {
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r = mid - 1;
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} else {
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l = mid + 1;
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}
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} while (l <= r);
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return false;
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}
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};
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```
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