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Leetcode 位运算 Single NumberII

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Single Number II

 Total Accepted: 14224 Total Submissions: 43648

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?


题意:在一组数组中除一个元素出现一次外其它元素都出现三次,找出这个元素

思路:位运算。因为数组中的元素是int类型的,所以可以开一个大小为sizeof(int) * 8的数组a来记录每一个bit出现的次数
最后对数组a中的每个元素中mod 3然后把它们组成一个int类型的数
这道题主要是位操作要熟练
写程序的时候,每一步都要想清楚,特别是数组的下标

复杂度:时间O(n),空间O(1)

相关题目:Single Number

class Solution {
public:
    int singleNumber(int A[], int n){
    	const int bits = sizeof(int) * 8;
    	int c[bits] = {0};
    	for(int i = 0; i < n; i++){
    		for(int j = 0; j < bits; j++){
    			c[j] += (A[i]>>j & 1);
    			c[j] %= 3;
    		}
    	}
    	int res = 0;
    	for(int i = 0; i < bits; i++){
    		res |= (c[i]<<i);
    	}
    	return res;
    }

};