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 in ascending from left to right.
- Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Given target = 5, return true.
Given target = 20, return false.
74. Search a 2D Matrix 的变形，这题的矩阵特点是：每一行是按从左到右升序排列；每一列从上到下按升序排列。
解法：有特点的数是左下角和右上角的数。比如左下角的18开始，上面的数比它小，右边的数比它大，和目标数相比较，如果目标数大，就往右搜，如果目标数小，就往上搜。这样就可以判断目标数是否存在。或者从右上角15开始，左面的数比它小，下面的数比它大。
Python:
class Solution:
def searchMatrix(self, matrix, target):
m = len(matrix)
if m == 0:
return False n = len(matrix[0])
if n == 0:
return False i, j = 0, n - 1
while i < m and j >= 0:
if matrix[i][j] == target:
return True
elif matrix[i][j] > target:
j -= 1
else:
i += 1 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 x = matrix.size() - 1, y = 0;
while (true) {
if (matrix[x][y] > target) --x;
else if (matrix[x][y] < target) ++y;
else return true;
if (x < 0 || y >= matrix[0].size()) break;
}
return false;
}
};
类似题目：
[LeetCode] 74. Search a 2D Matrix 搜索一个二维矩阵