最大子矩阵求解

最大子矩阵

1,借助最大连续子序列和来取得最大子矩阵

def MaxSubArray(nums):
    n = len(nums)
    dp = []
    dp.append(nums[0])
    mymax = dp[0]
    startidx = 0
    endidx = 0
    for i in range(1, n):
        if dp[i-1] + nums[i] <= nums[i] and dp[i-1] + nums[i] > 0:
            # print "startidx:%d" % i
            startidx = i
        dp.append(max(dp[i-1] + nums[i], nums[i]))
        if dp[i] > mymax:
            # print "endidx:%d" % i
            endidx = i
        mymax = max(dp[i], mymax)
    # print mymax
    return mymax, startidx, endidx

def MaxSubMatrix(mymatrix):
    row = len(mymatrix)
    col = len(mymatrix[0])
    # 循环外层
    #   固定外层i
    #      循环内层j
    #      找出该矩阵中最大的 子矩阵
    row_min,row_max,col_min,col_max = (0,0,0,0)
    mymax = 0
    for row_i in range(row):
        # print row_i
        for row_j in range(row_i, row):
            # print row_j
            numlist = []
            for col_i in range(col):
                # print col_i
                num = 0
                for row_tmp in range(row_i, row_j + 1):
                    num += mymatrix[row_tmp][col_i]
                numlist.append(num)
            tmpmax,startidx,endidx = MaxSubArray(numlist)
            if tmpmax > mymax:
                mymax = tmpmax
                row_min = row_i
                row_max = row_j
                col_min = startidx
                col_max = endidx
                print "mymax:%s,row_min:%s,row_max:%s,tartidx:%s,endidx:%s" % (mymax,row_min,row_max,startidx,endidx)

posted @ 2020-09-18 13:30  Leon呀  阅读(160)  评论(0编辑  收藏  举报