Single Number II LeetCode

Single Number II

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?

Solution

除了

every element appears two times except for one

参考这篇文章


除了某个数所有元素都出现2次

Bad Solution

这道题自己想不出来, 参考了别人的思路


本来自己想的是, 两个数字相同可以用异或来化为0


3个数字相同, 怎么化为0呢, 先乘后除, 先与后疑惑, 先加后减?

这些方案还要考虑独数在哪个位置, 乘除的先后顺序等, 不可行

Good Solution

网上看到的解决办法


int 一共有32位, 统计每位出现的次数. 独数的那一位出现的次数肯定是3n+1


可以把那一位出现次数取3的模, 左移 i 位


这儿有个技巧


本来是要分取3模是 0 还是 1 的, 但是0移位都是0


可以共用

res += (temp % 3) << i;

Code

[code]public class SingleNumberII {
    public static void main(String[] args) {
        System.out.println(singleNumber(new int[] { 1, 3, 3, 3 }));
        System.out.println(singleNumber(new int[] { 1, 4, 3, 4, 3, 4, 3 }));
        System.out.println(singleNumber(new int[] { 3, 3, 1, 3 }));
        System.out.println(singleNumber(new int[] { 3, 1, 3, 3 }));
        System.out.println(singleNumber(new int[] { 3, 1, 3, 3 }));
        System.out.println(singleNumber(new int[] { 3, 1, 4, 4, 3, 4, 3 }));

        System.out.println(singleNumber(new int[] { 5, 3, 2, 5, 4, 5, 4, 3, 4, 3 }));
        System.out.println(singleNumber(new int[] { 2, 2, 3, 2 }));

    }

    public static int singleNumber(int[] nums) {
        int temp = 0;
        int res = 0;
        for (int i = 0; i < 32; i++) {
            temp = 0;
            for (int j = 0; j < nums.length; j++) {
                temp  += (nums[j] >> i) & 1;
            }
            res += (temp % 3) << i;
        }

        return res;
    }

}

Note

这个是一个通用算法, 把 模3 改为 模k 就行