221 Maximum Square

这道题有两个思路， 一是沿用085的maximum rectangle的思路， 稍作改进即可， 代码如下， 这个方法运行192ms

class Solution:
    # @param {character[][]} matrix
    # @return {integer}
    def maximalSquare(self, matrix):
        ans = 0
        for i in range(0,len(matrix)):
            for j in range(0, len(matrix[0])):
                if i == 0:
                    matrix[i][j] = int(matrix[i][j])
                else:
                    if matrix[i][j] == "0":
                        matrix[i][j] = 0
                    else:
                        matrix[i][j] = matrix[i-1][j] + 1
        for m in matrix:
            ans = max(ans, self.largestRectangleArea(m))
        return ans

    def largestRectangleArea(self, height):
        ans = 0
        s = []
        for i in range(0, len(height)):
            left = i
            while (s != []) and (s[-1][0] > height[i]):
                left = s[-1][1]
                l = min(i - left, s[-1][0])
                ans = max(ans, l * l)
                s.pop()
            s.append([height[i], left])
        right = len(height)
        while s != []:
            left = s[-1][1]
            l = min(right - left, s[-1][0])
            ans = max(ans, l * l)
            s.pop()
        return ans

看到tag中有 dp, 试着用dp 做了一个 运行时间是200ms， 看出在复杂度都是O(n*n)的情况下 时间区别不大 代码如下

class Solution:
    # @param {character[][]} matrix
    # @return {integer}
    def maximalSquare(self, matrix):
        ans = 0
        m = len(matrix)
        if m == 0:
            return 0
        n = len(matrix[0])
        dp = [[0] * n for i in range(0,m)]
        for i in range(0,m):
            for j in range(0,n):
                dp[i][j] = int(matrix[i][j])
                if i and j and dp[i][j]:
                    dp[i][j] = 1 + min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1])
                    ans = max(ans, dp[i][j] * dp[i][j])
        return ans