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.
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
在一个给定的矩阵里查找某一个值，矩阵有如下特点：每一行里的整数都是按从左到右排序；每一行的第一数都大于前一行的最后一个数。
解法：二分法Binary Search，利用给定矩阵的特点，Z形大小顺序排列，把矩阵变成数组，使用二分法查找，或者直接在矩阵上使用二分法。
Python：
class Solution(object):
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
if not matrix:
return False
m, n = len(matrix), len(matrix[0])
left, right = 0, m * n
while left < right:
mid = left + (right - left) / 2
if matrix[mid / n][mid % n] >= target:
right = mid
else:
left = mid + 1
return left < m * n and matrix[left / n][left % n] == target
C++: Time: O(logm + logn)， Space: O(1)
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.empty()) {
return false;
}
// Treat matrix as 1D array.
const int m = matrix.size();
const int n = matrix[0].size();
int left = 0;
int right = m * n - 1;
// Find min of left s.t. matrix[left / n][left % n] >= target
while (left <= right) {
int mid = left + (right - left) / 2;
if (matrix[mid / n][mid % n] >= target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
// Check if matrix[left / n][left % n] equals to target.
if (left != m * n && matrix[left / n][left % n] == target) {
return true;
}
return false;
}
};
C++:
class Solution {
public:
bool searchMatrix(vector<vector<int> > &matrix, int target) {
if (matrix.empty() || matrix[0].empty()) return false;
if (target < matrix[0][0] || target > matrix.back().back()) return false;
int m = matrix.size(), n = matrix[0].size();
int left = 0, right = m * n - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (matrix[mid / n][mid % n] == target) return true;
else if (matrix[mid / n][mid % n] < target) left = mid + 1;
else right = mid - 1;
}
return false;
}
};