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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:
Example 1:
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true
Example 2:
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 100
-104 <= matrix[i][j], target <= 104
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int rows = matrix.size(),
cols = matrix[0].size(),
row = 0, col = cols - 1;
while (row < rows && col > -1) {
int cur = matrix[row][col];
if (cur == target) return true;
if (target > cur) row++;
else col--;
}
return false;
}
};
public boolean searchMatrix(int[][] matrix, int target) {
int i = 0, j = matrix[0].length - 1;
while (i < matrix.length && j >= 0) {
if (matrix[i][j] == target) {
return true;
} else if (matrix[i][j] > target) {
j--;
} else {
i++;
}
}
return false;
}
class Solution:
# @param matrix, a list of lists of integers
# @param target, an integer
# @return a boolean
# 8:21
def searchMatrix(self, matrix, target):
if not matrix or target is None:
return False
rows, cols = len(matrix), len(matrix[0])
low, high = 0, rows * cols - 1
while low <= high:
mid = (low + high) / 2
num = matrix[mid / cols][mid % cols]
if num == target:
return True
elif num < target:
low = mid + 1
else:
high = mid - 1
return False
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