解算almostIncreasingSequence（代码战）

Question

给定一个整数序列作为数组，通过从数组中删除不超过一个元素来确定是否可能获得严格递增的序列。

示例

对于序列[1, 3, 2, 1]，输出应该是：

almostIncreasingSequence(sequence) = false;

这个数组中没有一个元素可以删除以获得严格递增的序列。

对于序列[1, 3, 2]，输出应该是：

almostIncreasingSequence(sequence) = true.

您可以从数组中删除3以获得严格递增的序列[1，2]。或者，可以删除2以获得严格递增序列[1，3]。

我的代码：

def almostIncreasingSequence(sequence):
    c= 0
    for i in range(len(sequence)-1):
        if sequence[i]>=sequence[i+1]:
            c +=1
    return c<1

但它不能通过所有的测试。

input: [1, 3, 2]
Output:false
Expected Output:true

Input: [10, 1, 2, 3, 4, 5]
Output: false
Expected Output: true

Input: [0, -2, 5, 6]
Output: false
Expected Output: true

input:  [1, 1]
Output: false
Expected Output: true

Input: [1, 2, 3, 4, 3, 6]
Output: false
Expected Output: true

Input: [1, 2, 3, 4, 99, 5, 6]
Output: false
Expected Output: true

Answer 1

这是我的。希望你觉得这有帮助：

def almostIncreasingSequence(sequence):

    #Take out the edge cases
    if len(sequence) <= 2:
        return True

    #Set up a new function to see if it's increasing sequence
    def IncreasingSequence(test_sequence):
        if len(test_sequence) == 2:
            if test_sequence[0] < test_sequence[1]:
                return True
        else:
            for i in range(0, len(test_sequence)-1):
                if test_sequence[i] >= test_sequence[i+1]:
                    return False
                else:
                    pass
            return True

    for i in range (0, len(sequence) - 1):
        if sequence[i] >= sequence [i+1]:
            #Either remove the current one or the next one
            test_seq1 = sequence[:i] + sequence[i+1:]
            test_seq2 = sequence[:i+1] + sequence[i+2:]
            if IncreasingSequence(test_seq1) == True:
                return True
            elif IncreasingSequence(test_seq2) == True:
                return True
            else:
                return False