LeetCode_136. Single Number
2024-09-01 13:19:23
136. Single Number
Easy
Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,1]
Output: 1
Example 2:
Input: [4,1,2,1,2]
Output: 4
package leetcode.easy; import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Set; public class SingleNumber {
@org.junit.Test
public void test() {
int[] nums1 = { 2, 2, 1 };
int[] nums2 = { 4, 1, 2, 1, 2 };
System.out.println(singleNumber1(nums1));
System.out.println(singleNumber1(nums2));
System.out.println(singleNumber2(nums1));
System.out.println(singleNumber2(nums2));
System.out.println(singleNumber3(nums1));
System.out.println(singleNumber3(nums2));
System.out.println(singleNumber4(nums1));
System.out.println(singleNumber4(nums2));
} public int singleNumber1(int[] nums) {
HashSet<Integer> set = new HashSet<Integer>();
int result = 0;
for (int i = 0; i < nums.length; i++) {
if (!set.contains(nums[i])) {
set.add(nums[i]);
} else {
set.remove(nums[i]);
}
}
Iterator<Integer> it = set.iterator();
while (it.hasNext()) {
result = it.next();
}
return result;
} public int singleNumber2(int[] nums) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int result = 0;
for (int i = 0; i < nums.length; i++) {
if (!map.containsKey(nums[i])) {
map.put(nums[i], 1);
} else {
map.remove(nums[i]);
}
}
Set<Integer> set = map.keySet();
Iterator<Integer> it = set.iterator();
while (it.hasNext()) {
result = it.next();
}
return result;
} public int singleNumber3(int[] nums) {
HashSet<Integer> set = new HashSet<Integer>();
int sumSet = 0;
int sum = 0;
for (int i = 0; i < nums.length; i++) {
set.add(nums[i]);
sum += nums[i];
}
Iterator<Integer> it = set.iterator();
while (it.hasNext()) {
sumSet += it.next();
}
return 2 * sumSet - sum;
} public int singleNumber4(int[] nums) {
int a = 0;
for (int i = 0; i < nums.length; i++) {
a ^= nums[i];
}
return a;
}
}
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