题目描述 在一个 n * m 的二维数组中,每一行都按照从左到右递增的顺序排序,每一列都按照从上到下递增的顺序排序。请完成一个函数,输入这样的一个二维数组和一个整数,判断数组中是否含有该整数。
示例:
现有矩阵 matrix 如下:
[
[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]
]
给定 target = 5,返回 true。
给定 target = 20,返回 false。
限制:
0 <= n <= 10000 <= m <= 1000
解法 从左下角(或右上角)开始查找即可。 Python3
class Solution:
def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
if not matrix or not matrix[0]:
return False
rows, cols = len(matrix), len(matrix[0])
i, j = rows - 1, 0
while i >= 0 and j < cols:
if matrix[i][j] == target:
return True
if matrix[i][j] > target:
i -= 1
else:
j += 1
return False
Java
class Solution {
public boolean findNumberIn2DArray(int[][] matrix, int target) {
int m, n;
if (matrix == null || (m = matrix.length) == 0 || matrix[0] == null || (n = matrix[0].length) == 0) return false;
int i = 0, j = n - 1;
while (i < m && j >= 0) {
if (matrix[i][j] == target) return true;
if (matrix[i][j] > target) --j;
else ++i;
}
return false;
}
}
JavaScript
/**
* @param {number[][]} matrix
* @param {number} target
* @return {boolean}
*/
var findNumberIn2DArray = function (matrix, target) {
let m, n;
if (
matrix == null ||
(m = matrix.length) == 0 ||
matrix[0] == null ||
(n = matrix[0].length) == 0
)
return false;
let i = 0,
j = n - 1;
while (i < m && j >= 0) {
if (matrix[i][j] == target) return true;
if (matrix[i][j] > target) --j;
else ++i;
}
return false;
};
C++
class Solution {
public:
bool findNumberIn2DArray(vector<vector<int>>& matrix, int target) {
if (matrix.empty()) {
return false;
}
int m = matrix.size(), n = matrix[0].size();
int i = 0, j = n - 1;
while (i < m && j >= 0) {
if (matrix[i][j] == target) {
return true;
} else if (matrix[i][j] < target) {
++i;
} else {
--j;
}
}
return false;
}
};