leetcode1292

这道题暴力法会TLE，先给出这种TLE方案：

class Solution:
    def getMaxPoolVal(self,mat,length,m,n,threshold):
        maxval = 0
        for i in range(m-length+1):
            for j in range(n-length+1):
                curSquare = 0
                for x in range(i,i+length):#计算一个方形范围的所有值之和
                    for y in range(j,j+length):
                        curSquare += mat[x][y]
                if curSquare < threshold:
                    if curSquare > maxval: 
                        maxval = curSquare#更新最大值
                elif curSquare == threshold:
                    return threshold
        return maxval

    def maxSideLength(self, mat: 'List[List[int]]', threshold: int) -> int:
        m = len(mat)
        if m == 0:
            return 0
        n = len(mat[0])
        maxval = 0
        maxr = 0
        r = min(m,n)
        for t in range(r,0,-1):
            maxpool = self.getMaxPoolVal(mat,t,m,n,threshold)
            if maxpool >= maxval:
                maxval = maxpool
            if t >= maxr and maxpool != 0:
                maxr = t
            if maxval == threshold:
                return t
        return maxr

 

使用前序和方式，Java的可以AC，但是python仍然会TLE，给出这种TLE方案：

class Solution:
    def getMaxPoolVal(self,mat,length,m,n,threshold):
        for i in range(m-length+1):
            for j in range(n-length+1):
                curSquare = mat[i+length][j+length] - mat[i][j+length] - mat[i+length][j] + mat[i][j]#用累加矩阵降低计算量
                if curSquare <= threshold:
                    return True
        return False

    def maxSideLength(self, mat: 'List[List[int]]', threshold: int) -> int:
        m = len(mat)
        if m == 0:
            return 0
        n = len(mat[0])

        preSum = [[0 for _ in range(n+1)]for _ in range(m+1)]
        for i in range(1,m+1):#生成累加矩阵
            sums = 0
            for j in range(1,n+1):
                sums += mat[i-1][j-1]
                preSum[i][j] = preSum[i-1][j] + sums
        #print(preSum)
        r = min(m,n)
        for t in range(r,0,-1):#普通循环
            if self.getMaxPoolVal(preSum,t,m,n,threshold):
                return t
        return 0

参考：https://leetcode.com/problems/maximum-side-length-of-a-square-with-sum-less-than-or-equal-to-threshold/discuss/451843/Java-PrefixSum-solution

python要想AC，或者使用DP或者使用二分搜索：

最后给出AC的解决方案，将上面的24行的循环，改为二分查找：

class Solution:
    def getMaxPoolVal(self,mat,length,m,n,threshold):
        for i in range(m-length+1):
            for j in range(n-length+1):
                curSquare = mat[i+length][j+length] - mat[i][j+length] - mat[i+length][j] + mat[i][j]
                if curSquare <= threshold:
                    return True
        return False

    def maxSideLength(self, mat: 'List[List[int]]', threshold: int) -> int:
        m = len(mat)
        if m == 0:
            return 0
        n = len(mat[0])

        preSum = [[0 for _ in range(n+1)]for _ in range(m+1)]
        for i in range(1,m+1):
            sums = 0
            for j in range(1,n+1):
                sums += mat[i-1][j-1]
                preSum[i][j] = preSum[i-1][j] + sums
        #print(preSum)
        r = min(m,n)
        # for t in range(r,0,-1):
        #     if self.getMaxPoolVal(preSum,t,m,n,threshold):
        #         return t
        l,h = 0,r#改进为二分查找
        while l <= h:
            mid = l + (h - l) // 2
            if self.getMaxPoolVal(preSum,mid,m,n,threshold):
                l = mid + 1
            else:
                h = mid - 1
        return h

 

下面给出参考的方案：

二分搜索方案：https://leetcode.com/problems/maximum-side-length-of-a-square-with-sum-less-than-or-equal-to-threshold/discuss/451942/python-O(m*n*log(min(m-n)))

dp方案：https://leetcode.com/problems/maximum-side-length-of-a-square-with-sum-less-than-or-equal-to-threshold/discuss/451927/python-dp-accept

与本题相近的Java方案：https://leetcode.com/problems/maximum-side-length-of-a-square-with-sum-less-than-or-equal-to-threshold/discuss/451871/Java-sum%2Bbinary-O(m*n*log(min(mn)))

总结：这道题对我来说难度比较大，我只能作出TLE的解。