(LeetCode)Contains Duplicate II --- 查找重复的元素升级版
2016-07-27 10:23
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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 difference
between i and j is at most k.
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题意:
首先要读懂题意,题目意思理解了就简单了。
首先给出一个整数数组,和一个K元素
然后判断是否含有重复的元素,并比较两个重复元素的索引,下表 i, j 是否满足 | i - j | < K
如果满足返回True,否则返回 False。注意:最多只能有一对重复的。
解题分析:
首先判断是否有重复的元素,如果有找到两个元素的索引,此处可以用hash的方法,找到第一个的时候index保存它的索引,
再找到第二个的时候用循环的变量 i 保存第二个索引,然后进行 i - index 求出结果,返回True或者False.
between i and j is at most k.
Subscribe to see which companies asked this question
题意:
首先要读懂题意,题目意思理解了就简单了。
首先给出一个整数数组,和一个K元素
然后判断是否含有重复的元素,并比较两个重复元素的索引,下表 i, j 是否满足 | i - j | < K
如果满足返回True,否则返回 False。注意:最多只能有一对重复的。
解题分析:
首先判断是否有重复的元素,如果有找到两个元素的索引,此处可以用hash的方法,找到第一个的时候index保存它的索引,
再找到第二个的时候用循环的变量 i 保存第二个索引,然后进行 i - index 求出结果,返回True或者False.
# -*- coding:utf-8 -*- __author__ = 'jiuzhang' class Solution(object): def containsNearbyDuplicate(self, nums, k): numsDict = dict() for i in range(len(nums)): index = numsDict.get(nums[i]) if (index >= 0 and i - index <= k): return True numsDict[nums[i]] = i return False
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