LeetCode Single Number II

最新推荐文章于 2025-09-06 19:09:11 发布

转载最新推荐文章于 2025-09-06 19:09:11 发布 · 55 阅读

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原文链接：http://www.cnblogs.com/birdhack/p/3954418.html

文章标签：

#runtime

本文介绍了一种算法，用于在给定整数数组中找到仅出现一次的元素，所有其他元素都出现三次。算法具有线性时间复杂度且不使用额外内存。

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

public class Solution {
    public int singleNumber(int[] A) {
            HashMap<Integer, Integer> map = new HashMap<>();
            int singnum=0;
            for (int i = 0; i < A.length; i++) {
                if (map.containsKey(A[i])) {
                    map.put(A[i], map.get(A[i])+1);
                }else {
                    map.put(A[i],1);
                }
                if (map.get(A[i])==3) {
                    map.remove(A[i]);
                }
            }
            for (Integer i : map.keySet()) {
                singnum=i;
            }
            return singnum;
    }
}