Single Number II - leetcode

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Single Number II

AC Rate: 11/62

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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?

思路是：利用一个33位的数组保存n个数的每个二进制位1的个数，然后对其模3 ，这样最后剩下的就是出现一次的那个数了，注意第32位保存的正负数的个数，同样也要模3，最后决定结果的正负，同时要考虑溢出的情况。

代码如下

class Solution {public:    int singleNumber(int A[], int n) {        // Note: The Solution object is instantiated only once and is reused by each test case.        if ( n < 4)            return A[0];        short tmp[33] = {0};        for (int i = 0; i < n; ++i) {            unsigned int t = 0;            if (A[i] < 0) {                t = -A[i];                tmp[32] += 1;                tmp[32] %= 3;            }            else {                t = A[i];            }            int j = 0;            while (t) {                tmp[j] += t & 0x01;               // cout << (t & 0x01) << " ";                tmp[j] %= 3;                t >>= 1;                ++j;            }        }        unsigned int key = 0;        for (int i = 0; i < 32; ++i) {            if (tmp[i])                key += (1 << i);        }        return tmp[32] == 0 ? key : -key;    }};