Question

给定X个整数和整数Q的P，检查P是否包含数字1,2，...，Q。

例如，给定以下数组P和Q = 3

P[0] = 1
P[1] = 1
P[2] = 2
P[3] = 3
P[4] = 3

该函数应返回True。

以下是我的所得，但我不确定我是否正确？

def solution(P, Q):
    n = len(P)
    for i in xrange(n - 1):
        if (P[i] + 1 < P[i + 1]):
            return False
    if (P[0] != 1 and P[n - 1] != Q):
        return False
    else:
        return True

Answer 1

将数组转换为集合，并将集合与1到K之间的所有整数集合进行比较：

// Delegate submit action
$(document).on('submit', 'form', function () {

    var validate = true;
    var unanswered = new Array();

    // Loop through available sets
    $('.option').each(function () {
        // Question text
        var question = $(this).prev();
        // Validate
        if (!$(this).find('input').is(':checked')) {
            // Didn't validate ... dispaly alert or do something
            unanswered.push(question.text());
            question.css('color', 'red'); // Highlight unanswered question
            validate = false;
        }
    });

    if (unanswered.length > 0) {
        msg = "Please answer the following questions:\n" + unanswered.join('\n'); 
        alert(msg);
    }
    return validate;
});

作为额外的奖励，set(A)==set(range(1,K+1))不再需要排序。

Answer 2

当我理解你的问题时，数组需要增加1或从i增加到i + 1。因此，您需要更改if语句中的符号。

def solution(A, K):
    n = len(A)
    for i in xrange(n - 1):
        if (A[i] != A[i + 1] and A[i] + 1 != A[i + 1]):
            return False
    if (A[0] != 1 or A[n - 1] != K):
        return False
    else:
        return True

>>>solution([1,2,3],3)
True
>>>solution([1,2,2,3,3],3)
True
>>>solution([1,2,3],4)
False
>>>solution([1,2,3],2)
False

Answer 3

def solution(A, K):
n = len(A)
for i in range(K-1):
    if (A[i] + 1 < A[i + 1]):
        return False
if (A[0] != 1 or A[K - 1] != K):
    return False
else:
    return True

检查数组是否包含特定的数字列表

3 个答案: