LeetCode 217. Contains Duplicate

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.

 

分析:

HashSet?

 

first try:

import java.util.HashSet;

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

 

 

result:

 

second:

import java.util.HashSet;

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

 

 

result:

the same.

 

third:

import java.util.HashSet;

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

 

result:

 

anylysis:

 

看了solution里的代码,并执行了一下,发现比我的代码还慢。

import java.util.HashSet;

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

 

result:

32.96%

import java.util.HashSet;

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

 

17.88%

发现慢的原因在于Set,如果是HashSet速度会变快。

 

for 和foreach循环速度在都是HashSet时没有变化;在是Set时有变化。

 

发现这种优化费时费力,放弃。

 

 

 

总结:

import java.util.HashSet;

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

posted @ 2018-04-19 14:34  Zhao_Gang  阅读(85)  评论(0编辑  收藏  举报