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algorithmsbypython's Issues

请教一下二位数组查找

二维数组查找的类型判断这部分代码想请教一下,没看懂

 if type(target) == float and type(array[0][0]) == int:
    if int(target) == target:
        return False
    target = int(target)
    elif type(target) == int and type(array[0][0]) == float:
        target = float(int)
    elif type(target) != type(array[0][0]):     # 浮点数的相等判断问题需要特别注意, 一般都是判断两个数的差值是否小于一个特别小的数。这里不展开分析。
        return False

剑指offer+面试题3+第31行+是不是有问题

    if type(target) == float and type(array[0][0]) == int:
        if int(target) == target:      #这里的==应该是!=吧 第三个测试例子也因为这个==,
                                                  #导致13.0输出为False,但是明明array里有13
              return False

旋转数组三个值一样的情况,应该直接return吧?

            elif rotateArray[mid] == rotateArray[front] and rotateArray[front] == rotateArray[rear]:
                for i in range(1, len(rotateArray)):
                    if rotateArray[i] < minVal:
                        minVal = rotateArray[i]
                        rear = i

这一段代码,mid,front,rear的值都一样之后,遍历整个数组找出最小值了,应该就直接return吧,怎么又进循环了。这里又多循环几次,切到只剩两个数了才出循环,不是做白工了吗。第一次就已经在整个list里遍历了一遍,找出最小值了,再循环没意义啊。

旋转数组的最小值,下面算法和lz的一样?

说实话我没看明白lz写的,但是测试了下我的,居然一样

class Solution:
    def minNumberInRotateArray(self, rotateArray):
        if len(rotateArray)==0:
            return 0
        head =0
        rear = len(rotateArray)-1
        while rotateArray[head]<=rotateArray[head+1]:
            head+=1
        return rotateArray[head+1]

targetofferbyPython 稀疏矩阵转置、丑数优化

def transTriple(triple):
m, n = shape(triple)
transMatrix = []
# sortedIndex = array([m[1] for m in triple]).argsort()
for i in range(m):
# tempArray = triple[sortedIndex[i]]
tempArray = triple[i]
transMatrix.append([tempArray[1], tempArray[0], tempArray[2]])
return transMatrix

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