LeetCode:Single Number II
2014-05-19 23:14
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题目:
Given an array of integers, every element appearsthree 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?
解题思路:
思路类似基数排序,将所有数字列成n行,然后观察其中的一列,你会发现,由于其他数字都出现了
三次,如果当前这列某个数位的个数不能被3整除,那么,所要求的数在此位必等于这个数位的值,这样
就能以O(n)的时间复杂度解决问题了(注:此处常系数是10).
下面是解题代码.
注:abs((long long)A[j])中要将A[j]强制转换为long long型,不然程序有bug,比如A[j]等于-2^31.
Given an array of integers, every element appearsthree 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?
解题思路:
思路类似基数排序,将所有数字列成n行,然后观察其中的一列,你会发现,由于其他数字都出现了
三次,如果当前这列某个数位的个数不能被3整除,那么,所要求的数在此位必等于这个数位的值,这样
就能以O(n)的时间复杂度解决问题了(注:此处常系数是10).
下面是解题代码.
class Solution { public: int singleNumber(int A[], int n) { int div[10] = {1} , cnt = 0 , res = 0; for(int i=1;i<10;++i) div[i] = div[i-1] * 10 ; for(int i=0;i<n;++i) cnt +=A[i] < 0 ; for(int i=1;i<=10;++i) { int hash[10] = { 0 } ; for(int j=0;j<n;++j) ++hash[abs((long long)A[j])/div[i-1]%10]; for(int j=0;j<10;++j) if(hash[j] % 3) res = j * div[i-1] + res ; } return cnt % 3 ? -res : res ; } };
注:abs((long long)A[j])中要将A[j]强制转换为long long型,不然程序有bug,比如A[j]等于-2^31.
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