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## Problem

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. 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.

## Code

class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int row = 0;
int col = matrix[0].length - 1;

while(col >= 0 && row < matrix.length) {
int num = matrix[row][col];
if(num == target) return true;

if(num > target) {
col--;
} else {
row++;
}
}

return false;
}
}

class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""

if not matrix or matrix and not matrix[0]:
return False

# 从右上角搜索
row = 0
col = len(matrix[0]) - 1

while col >= 0 and row <= len(matrix) - 1:
num = matrix[row][col]
if target == num:
return False
elif target < num:
col-= 1
else:
row += 1

return False