本问题对应的 leetcode 原文链接：剑指 Offer 56 – II. 数组中数字出现的次数 II
问题描述
在一个数组 nums 中除一个数字只出现一次之外，其他数字都出现了三次。请找出那个只出现一次的数字。
示例 1:
输入：nums = [3,4,3,3]
输出：4
示例 2:
输入：nums = [9,1,7,9,7,9,7]
输出：1
限制：
1 <= nums.length <= 100001 <= nums[i] < 2^31
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
public int singleNumber(int[] nums) {
int[] res = new int[32];
int m = 1;
int sum = 0;
for(int i = 0; i < 32; i++){
for(int j = 0; j < nums.length; j++){
if((nums[j] & m) != 0){
res[i]++;
}
}
res[i] = res[i] % 3;
sum = sum + res[i] * m;
m = m << 1;
}
return sum;
}
}
时间复杂度：O(n)
额外空间复杂度：O(1)
Python
class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
res = [0] * 32
m = 1
sum = 0
for i in range(32):
for j in range(len(nums)):
if (nums[j] & m) != 0:
res[i] += 1
res[i] %= 3
sum += res[i] * m
m <<= 1
return sum
C++
class Solution {
public:
int singleNumber(vector<int>& nums) {
vector<int> res(32, 0);
unsigned int m = 1;
int sum = 0;
for (int i = 0; i < 32; i++) {
for (int j = 0; j < nums.size(); j++) {
if ((nums[j] & m) != 0) {
res[i]++;
}
}
res[i] %= 3;
sum += res[i] * m;
m <<= 1;
}
return sum;
}
};
Go
func singleNumber(nums []int) int {
res := make([]int, 32)
m := 1
sum := 0
for i := 0; i < 32; i++ {
for j := 0; j < len(nums); j++ {
if (nums[j] & m) != 0 {
res[i]++
}
}
res[i] %= 3
sum += res[i] * m
m <<= 1
}
return sum
}
JS
/**
* @param {number[]} nums
* @return {number}
*/
var singleNumber = function(nums) {
let res = new Array(32).fill(0);
let m = 1;
let sum = 0;
for (let i = 0; i < 32; i++) {
for (let j = 0; j < nums.length; j++) {
if ((nums[j] & m) != 0) {
res[i]++;
}
}
res[i] = res[i] % 3;
sum = sum + res[i] * m;
m = m << 1;
}
return sum;
};
shuaidi