README_EN.md

217. Contains Duplicate

中文文档

Description

Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

 

Example 1:

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

Example 2:

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

Example 3:

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

 

Constraints:

• 1 <= nums.length <= 105
• -109 <= nums[i] <= 109

Solutions

Python3

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        return len(nums) != len(set(nums))

Java

class Solution {
    public boolean containsDuplicate(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int num : nums) {
            if (s.contains(num)) {
                return true;
            }
            s.add(num);
        }
        return false;
    }
}

TypeScript

function containsDuplicate(nums: number[]): boolean {
    let unique: Set<number> = new Set(nums);
    return unique.size != nums.length;
}

C++

class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_set<int> s;
        for (int e : nums) {
            if (s.count(e)) return true;
            s.insert(e);
        }
        return false;
    }
};

Go

func containsDuplicate(nums []int) bool {
	s := make(map[int]bool)
	for _, e := range nums {
		if s[e] {
			return true
		}
		s[e] = true
	}
	return false
}

C#

public class Solution {
    public bool ContainsDuplicate(int[] nums) {
        return nums.Distinct().Count() < nums.Length;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var containsDuplicate = function (nums) {
    return new Set(nums).size !== nums.length;
};

Rust

use std::collections::HashSet;
impl Solution {
    pub fn contains_duplicate(nums: Vec<i32>) -> bool {
        nums.iter().collect::<HashSet<&i32>>().len() != nums.len()
    }
}

impl Solution {
    pub fn contains_duplicate(mut nums: Vec<i32>) -> bool {
        nums.sort();
        let n = nums.len();
        for i in 1..n {
            if nums[i - 1] == nums[i] {
                return true
            }
        }
        false
    }
}

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