Given a 2D binary matrix filled with 0‘s and 1‘s,find the largest square containing only 1‘s and return its area.
Example:
Input: 1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0 Output: 4
思路是DP,3种做法,通用的T: O(m*n),S: O(m*n) 和只针对部分情况可以use 滚动数组来reduce space成为O(n).
A[i][j] = min(A[i-1][j-1],left[i][j-1],up[i-1][j]) + 1 为边长 i,j > 0
滚动数组
A[i][j] = min(A[i-1][j-1],A[i][j-1],A[i-1][j]) + 1 为边长 i,j > 0
A[i][j] = min(A[i%2-1][j-1],A[i%2][j-1],A[i%2-1][j]) + 1 为边长 i,j > 0
1. Constraints
1) size >=[0*0]
2) element will be "1" or "0" # note it will be integer or string
2. Ideas
DP T: O(m*n) S: O(n) optimal
1) edge case,empty,m == 1 or n == 1
2) left,up,ans,init
3)
A[i][j] = min(A[i-1][j-1],up[i-1][j]) + 1
4) return res*res
3. codes
1) use left,and ans T: O(m*n) S: O(m*n)
1 class Solution: 2 def maxSquare(self,matrix): 3 if not matrix: return 0 4 m,n = len(matrix),len(matrix[0]) 5 left,res = [[0]*n for _ in range(m)],[[0]*n for _ in range(m)],[[0]*n for _ in range(m)],0 6 for i in range(m): 7 for j in range(n): 8 if matrix[i][j] == "1": 9 res = 1 # edge case when m == 1 or n == 1 10 if j == 0: 11 left[i][j] = ans[i][j] = 1 12 if i == 0: 13 up[i][j] = ans[i][j] = 1 14 if i >0 and j > 0: 15 left[i][j] = left[i][j-1] + 1 16 up[i][j] = up[i-1][j] + 1 17 for i in range(1,m): 18 for j in range(1,n): 19 if matrix[i][j] == "1": 20 ans[i][j] = min(ans[i-1][j-1],up[i-1][j])+1 21 res = max(res,ans[i][j]) 22 return res*res
3.2) skip left and up,just use ans array
1 class Solution: 2 def maxSquare(self,len(matrix[0]) 5 ans,temp,res = [[0]*n for _ in range(m)],0 6 for i in range(m): 7 for j in range(n): 8 if matrix[i][j] == "1": 9 temp = 1 10 ans[i][0] = int(matrix[i][0]) 11 ans[0][j] = int(matrix[0][j]) 12 if i > 0 and j >0 : 13 ans[i][j] = min(ans[i-1][j-1],ans[i][j-1],ans[i-1][j]) + 1 14 res = max(res,ans[i][j]) 15 16 return max(res,temp )**2
3.3) 滚动数组, T: O(m*n), S: O(n)
1 class Solution: 2 def maxSquare(self,res = [[0]*n for _ in range(2)],0 6 for j in range(n): # note initialize when using rolling array 7 if matrix[0][j] == "1": 8 temp = 1 9 ans[0][j] = int(matrix[0][j]) 10 for i in range(1,m): 11 for j in range(n): 12 if j == 0: # initialize 13 ans[i%2][0] = int(matrix[i][0]) 14 if matrix[i][j] == "1": 15 temp = 1 16 if i > 0 and j >0 : 17 ans[i%2][j] = min(ans[i%2-1][j-1],ans[i%2][j-1],ans[i%2-1][j]) + 1 18 res = max(res,ans[i%2][j]) 19 else: # very important for initialize 20 ans[i%2][j] = 0 21 return max(res,temp )**2
4. Test cases
1) edge case
2)
1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0
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