题目地址

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

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

``````class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.empty() || matrix[0].empty()) return false;
int left = 0, right = matrix.size();
while (left < right) {
int mid = (left + right) / 2;
if (matrix[mid][0] == target) return true;
if (matrix[mid][0] <= target) left = mid + 1;
else right = mid;
}
int tmp = (right > 0) ? (right - 1) : right;
left = 0;
right = matrix[tmp].size();
while (left < right) {
int mid = (left + right) / 2;
if (matrix[tmp][mid] == target) return true;
if (matrix[tmp][mid] < target) left = mid + 1;
else right = mid;
}
return false;
}
};

``````

``````class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.empty() || matrix[0].empty()) return false;
int m = matrix.size(), n = matrix[0].size();
int left = 0, right = m * n;
while (left < right) {
int mid = (left + right) / 2;
if (matrix[mid / n][mid % n] == target) return true;
if (matrix[mid / n][mid % n] < target) left = mid + 1;
else right = mid;
}
return false;
}
};

``````

``````class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.empty() || matrix[0].empty()) return false;
int i = 0, j = (int)matrix[0].size() - 1;
while (i < matrix.size() && j >= 0) {
if (matrix[i][j] == target) return true;
else if (matrix[i][j] > target) --j;
else ++i;
}
return false;
}
};
``````

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