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Single Number II
Given an array of integers, every element appears threetimes except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
解法一：开辟map记录次数
class Solution { public: int singleNumber(int A[], int n) { map<int, int> m; for(int i = 0; i < n; i ++) { map<int, int>::iterator it = m.find(A[i]); if(it == m.end()) m[A[i]] = 1; else m[A[i]] ++; } for(map<int, int>::iterator it = m.begin(); it != m.end(); it ++) { if(it->second != 3) return it->first; } } };
解法二：先排序，再遍历找出孤异元素
class Solution { public: int singleNumber(int A[], int n) { sort(A, A+n); //note that at least 4 elements in A if(A[0] != A[1]) return A[0]; if(A[n-1] != A[n-2]) return A[n-1]; for(int i = 1; i < n-1; i ++) { if(A[i] != A[i-1] && A[i] != A[i+1]) return A[i]; } } };
解法三：位操作。不管非孤异元素重复多少次，这是通用做法。
对于右数第i位，如果孤异元素该位为0，则该位为1的元素总数为3的整数倍。
如果孤异元素该位为1，则该位为1的元素总数不为3的整数倍（也就是余1）。
换句话说，如果第i位为1的元素总数不为3的整数倍，则孤异数的第i位为1，否则为0.
（如果非孤异元素重复n次，则判断是否为n的整数倍）
class Solution { public: int singleNumber(int A[], int n) { int count; int ind = 1; //mask position int result = 0; while(ind) { count = 0; for(int i = 0; i < n; i ++) { if(A[i] & ind) count ++; } if(count % 3) result |= ind; //position ind is 1 ind <<= 1; } return result; } };
【LeetCode】Single Number II (3 solutions)
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原文地址：http://www.cnblogs.com/ganganloveu/p/4110996.html
