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.
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
public class ContainsDuplicate {
public static void main(String[] args) {
int [] nums = {3,3};
System.out.println(containsDuplicate4(nums));
}
//最蠢的方法,超时了
public static boolean containsDuplicate(int[] nums) {
boolean flag = false;
for (int i = 0; i < nums.length; i++) {
for (int j = i+1; j < nums.length; j++) {
if (nums[i] == nums[j]) {
flag = true;
}
}
}
return flag;
}
//把元素存储到hashmap中,元素为key,如果该映射以前包含了一个该键的映射关系,则旧值被替换
//所以一旦重复那么hashmap中的元素个数必然小于实际存入的个数
//用时 20ms
public static boolean containsDuplicate2(int[] nums) {
boolean flag = false;
Map map = new HashMap();
for (int i = 0; i < nums.length; i++) {
map.put(nums[i], i);
if (map.size() map = new HashMap();
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(nums[i])) {
flag = true;
break;
}else {
map.put(nums[i], i);
}
}
return flag;
}
//使用set
//用时 15ms
public static boolean containsDuplicate4(int[] nums) {
Set set = new HashSet(nums.length);
for (int i = 0; i < nums.length; i++) {
if (set.contains(nums[i])) {
return true;
}else {
set.add(nums[i]);
}
}
return false;
}
//使用ArrayList
//Time Limit Exceeded (为啥超时啦)
public static boolean containsDuplicate5(int[] nums) {
List list = new ArrayList();
for (int i = 0; i < nums.length; i++) {
if (list.contains(nums[i])) {
return true;
}else {
list.add(nums[i]);
}
}
return false;
}
}
最后一个方法会超时,是因为ArrayList底层是数组实现的,如果要测试的数组元素特别多,那么可能遍历的就会多,而hashmap和hashset底层都是根据哈希值进行排序,重复的值在内存模型中是相邻的。所以遍历的次数少。(目前理解到这里。。。)