217. Contains Duplicate

于 2019-02-21 18:43:12 发布

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CC 4.0 BY-SA版权

文章标签： java runtime

原文链接：https://segmentfault.com/a/1190000018231266

本文介绍了一种使用HashSet解决数组中是否存在重复元素的问题的方法。通过遍历数组并将元素添加到HashSet中，若发现已存在元素则返回true，表示数组中存在重复。此方法简单高效，运行时间为10ms，内存使用为40.9MB。

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Given an array of integers, find if the array contains any duplicates.
Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.
Example 1:

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

Example 2:

Input: [1,2,3,4]
Output: false

Example 3:

Input: [1,1,1,3,3,4,3,2,4,2]
Output: true

难度：easy

题目：给定整数数组，找出其中是否含有重复元素。方法应该返回true如果发现某元素至少出现两次，否则返回false.

思路：hashset

Runtime: 10 ms, faster than 64.18% of Java online submissions for Contains Duplicate.
Memory Usage: 40.9 MB, less than 81.42% of Java online submissions for Contains Duplicate.

class Solution {
    public boolean containsDuplicate(int[] nums) {
        Set<Integer> set = new HashSet<>();
        for (int i = 0; i < nums.length; i++) {
            if (set.contains(nums[i])) {
                return true;
            }
            set.add(nums[i]);
        }
        
        return false;
    }
}