LeetCode--219、268、283、414、448 Array(Easy)

 

219. Contains Duplicate II

Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

题目地址：https://leetcode.com/problems/contains-duplicate-ii/description/

题意：给定一个数组和一个整数k，判断是否存在两个相等数的下标之差绝对值为k

思路：hash

class Solution(object):
    def containsNearbyDuplicate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        d = {}
        for i, n in enumerate(nums):
            if n in d:
                if abs(i - d[n]) <= k:
                    return True
            d[n] = i
        return False

 

268. Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example 1

Input: [3,0,1]
Output: 2

Example 2

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

题目地址：https://leetcode.com/problems/missing-number/description/

题意：找0~n中缺少的数

思路：首项为0公差为1的等差数列和，减去数组的和

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        return n * (n+1) / 2 - sum(nums)

 

283. Move Zeroes

Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.

For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].

Note:

1. You must do this in-place without making a copy of the array.
2. Minimize the total number of operations.

题目地址：https://leetcode.com/problems/move-zeroes/description/

题意：将数组的0移到数组的后面，其他元素的相对顺序不变

思路：遍历，为0则删除加到末尾

class Solution(object):
    def moveZeroes(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        for n in nums:
            if n == 0:
                nums.remove(n)
                nums.append(n)

 

414. Third Maximum Number

Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

Example 1:

Input: [3, 2, 1]
Output: 1
Explanation: The third maximum is 1.

Example 2:

Input: [1, 2]
Output: 2
Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: [2, 2, 3, 1]
Output: 1
Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.

题目地址：https://leetcode.com/problems/third-maximum-number/description/

题意：返回数组中第三大的数

思路：使用set使数组元素唯一，排序输出

class Solution(object):
    def thirdMax(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return sorted(set(nums))[-3] if len(set(nums))>2 else max(nums)

 

448. Find All Numbers Disappeared in an Array

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

Input:
[4,3,2,7,8,2,3,1]
Output:
[5,6]

题目地址：https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/

题意：给定长度n的数组，找出[1,n]不在数组中的元素

思路：

class Solution(object):
    def findDisappearedNumbers(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        for i in xrange(len(nums)):
            index = abs(nums[i]) - 1
            nums[index] = - abs(nums[index])
        return [i + 1 for i in range(len(nums)) if nums[i] > 0]