代码挑战:找到单一数字-Java
【摘要】 Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Given a non-empty array of integers, every element appears twice except for one. Find that single one.
方案一:
class Solution {
public int singleNumber(int[] nums) {
int result = nums[0];
for (int i = 1; i < nums.length; i++) {
result = result ^ nums[i];
}
return result;
}
}
方案二:
class Solution {
public int singleNumber(int[] nums) {
Set<Integer> set = new HashSet<Integer>();
for (int i = 0; i < nums.length; i++) {
Integer v = nums[i];
if(set.contains(v)) {
set.remove(v);
} else {
set.add(v);
}
}
return Integer.parseInt(set.toArray()[0].toString());
}
}
测试案例:
@Test
public void test1() {
Solution solution = new Solution();
int[] nums = new int[] { 2, 2, 1 };
int result = solution.singleNumber(nums);
Assert.assertEquals(result, 1);
}
@Test
public void test2() {
Solution solution = new Solution();
int[] nums = new int[] { 4,1,2,1,2 };
int result = solution.singleNumber(nums);
Assert.assertEquals(result, 4);
}
【声明】本内容来自华为云开发者社区博主,不代表华为云及华为云开发者社区的观点和立场。转载时必须标注文章的来源(华为云社区)、文章链接、文章作者等基本信息,否则作者和本社区有权追究责任。如果您发现本社区中有涉嫌抄袭的内容,欢迎发送邮件进行举报,并提供相关证据,一经查实,本社区将立刻删除涉嫌侵权内容,举报邮箱:
cloudbbs@huaweicloud.com
- 点赞
- 收藏
- 关注作者
评论(0)