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Copy path136_Single_Number.java
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139 lines (124 loc) · 3.4 KB
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/*
* 136. Single Number
* Target: Given a non-empty array of integers, every element appears twice except for one. Find that single one.
* Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
* Difficulty：Easy
* Classification：HashTable, Bit Manipulation
* Best solution: sol 5.
*/
/*
* Solution 1
* 2019-07-26 Runtime: 305 ms
* Algorithm: => Brutal. List operation
* Time Complexity: O(n^2). Space Conplexity: O(n)
*/
class Solution {
public int singleNumber(int[] nums) {
if (nums.length == 0) return 0;
List<Integer> res = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
if (!res.contains(nums[i])) {
res.add(nums[i]);
} else {
for (int j = 0; j < res.size(); j++) {
if (nums[i] == res.get(j)) {
res.remove(j);
}
}
}
}
return res.get(0);
}
}
/*
* Solution 2
* 2019-07-26 Runtime: 9 ms
* Algorithm: => HashTable. If not contains, add it. Otherwise, remove it. The rest number is what we want.
* Time Complexity: O(n). Space Conplexity: O(n)
*/
import java.util.Enumeration;
import java.util.Hashtable;
class Solution {
public int singleNumber(int[] nums) {
if (nums.length == 0) return 0;
Hashtable<Integer, Integer> res = new Hashtable<>();
for (int i = 0; i < nums.length; i++) {
if (!res.containsKey(nums[i])) {
res.put(nums[i], 1);
} else {
res.remove(nums[i]);
res.put(nums[i], 2);
}
}
Enumeration e = res.keys();
while( e.hasMoreElements() ){
Object x = e.nextElement();
if(res.get(x)==1){
return (int)x;
}
}
return 0;
}
}
/*
* Solution 3
* 2019-07-26 Runtime: 7 ms
* Algorithm: => HashSet. Same as sol 2.
* Time Complexity: O(n). Space Conplexity: O(n)
*/
class Solution {
public int singleNumber(int[] nums) {
Set<Integer> s = new HashSet<>();
for (int num : nums) {
if (!s.add(num)) {
s.remove(num);
}
}
return s.iterator().next();
}
}
/*
* Solution 4
* 2019-07-26 Runtime: 6 ms
* Algorithm: => Math and HashSet. Based on formula: 2∗(a+b+c) − (a+a+b+b+c) = c.
* Time Complexity: O(n). Space Conplexity: O(n)
*/
class Solution {
public int singleNumber(int[] nums) {
Set<Integer> s = new HashSet<>();
for (int num : nums) {
s.add(num);
}
return 2 * sum2(s) - sum1(nums);
}
private int sum1(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
private int sum2(Set<Integer> s) {
int sum = 0;
for (int str : s) {
sum += str;
}
return sum;
}
}
/*
* Solution 5
* 2019-07-26 Runtime: 0 ms
* Algorithm: => Math: XOR. a⊕b⊕a = (a⊕a)⊕b = 0⊕b = b
* So we can XOR all bits together to find the unique number.
* Time Complexity: O(n). Space Conplexity: O(1)
*/
class Solution {
public int singleNumber(int[] nums) {
int res = 0;
for (int num : nums) {
res ^= num;
}
return res;
}
}