[LeetCode] Single Number II, Solution
2013-11-18 07:01
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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? [Thoughts] 这个题有类似于single number的解法,即通过位运算,一遍扫描得到结果。还是读书的时候见过,大概是两个变量,相互做异或、补之类的运算,早不记得详情了。 现在的解法是比较普通的。因为题目已经说了,除了一个数字以外,其他的都出现了3次,如果我们把那个特殊的数剔除,并把剩下的数于每一位来加和的话,每一位上1出现的次数必然都是3的倍数。所以,解法就在这里,将每一位数字分解到32个bit上,统计每一个bit上1出现的次数。最后对于每一个bit上1出现的个数对3取模,剩下的就是结果。 [code] int singleNumber(int A[], int n) { vector<int> bit(32,0); for(int i =0; i< n; ++i) { int k=1; for(int j =0; j<32; ++j) { int rotated; if((rotated = A[i]>>j) == 0) break; bit[j] += rotated & k; } } int target=0; for(int i =0; i<32; ++i) { target += (bit[i]%3 <<i); } return target; }
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? [Thoughts] 这个题有类似于single number的解法,即通过位运算,一遍扫描得到结果。还是读书的时候见过,大概是两个变量,相互做异或、补之类的运算,早不记得详情了。 现在的解法是比较普通的。因为题目已经说了,除了一个数字以外,其他的都出现了3次,如果我们把那个特殊的数剔除,并把剩下的数于每一位来加和的话,每一位上1出现的次数必然都是3的倍数。所以,解法就在这里,将每一位数字分解到32个bit上,统计每一个bit上1出现的次数。最后对于每一个bit上1出现的个数对3取模,剩下的就是结果。 [code] int singleNumber(int A[], int n) { vector<int> bit(32,0); for(int i =0; i< n; ++i) { int k=1; for(int j =0; j<32; ++j) { int rotated; if((rotated = A[i]>>j) == 0) break; bit[j] += rotated & k; } } int target=0; for(int i =0; i<32; ++i) { target += (bit[i]%3 <<i); } return target; }
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