【leetcode】1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold
题目如下:
Given a
m x n
matrixmat
and an integerthreshold
. Return the maximum side-length of a square with a sum less than or equal tothreshold
or return 0 if there is no such square.Example 1:
Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4
Output: 2
Explanation: The maximum side length of square with sum less than 4 is 2 as shown.Example 2:
Input: mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1
Output: 0Example 3:
Input: mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6
Output: 3Example 4:
Input: mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184
Output: 2Constraints:
1 <= m, n <= 300
m == mat.length
n == mat[i].length
0 <= mat[i][j] <= 10000
0 <= threshold <= 10^5
解题思路:记grid[i][j]为左上顶点坐标是(0,0),右下顶点坐标为(i,j) 组成的矩形中每个坐标的和,grid很容易通过遍历一次mat求得。接下来就只要求mat中每个子矩形的和,如下图所示,要求绿色部分的矩形和,假设绿色矩形的右下顶点坐标是(i,j)只需要用grid[i][j] 减去两个黄色部分的矩形和,因为红色部分是两个黄色部分共有,因此做减法的时候多减了一次,需要再加回去,最终计算公式有:grid[i][j] - 两个黄色子矩形和 + 红色子矩形和。
代码如下:
class Solution(object):
def maxSideLength(self, mat, threshold):
"""
:type mat: List[List[int]]
:type threshold: int
:rtype: int
"""
grid = [[0] * len(mat[0]) for _ in mat]
for i in range(len(mat)):
amount = 0
for j in range(len(mat[i])):
amount += mat[i][j]
grid[i][j] = amount
if i > 0:
grid[i][j] += grid[i-1][j]
#print grid res = 0
for i in range(len(grid)):
for j in range(len(grid[i])): low,high = res,min(len(grid),len(grid[0]))
while low <= high:
mid = (low + high)/2
if i + mid >= len(grid) or j + mid >= len(grid[0]):
high = mid - 1
continue
val = grid[i + mid][j + mid]
if j - 1 >= 0:
val -= grid[i + mid][j - 1]
if i - 1 >= 0:
val -= grid[i - 1][j + mid]
if i - 1 >= 0 and j - 1 >= 0:
val += grid[i - 1][j - 1]
if val <= threshold:
res = max(res, mid + 1)
low = mid + 1
else:
high = mid - 1 return res
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