算法第十八周作业01

Description

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up:

Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)

Solution

采用Map结构存储每个元素及其出现的次数

逐个判断Map中的每个元素，如果其出现次数为1，则记录到输出结果集里面

Code

class Solution {
public:
map<int, int> m;
vector<int> singleNumber(vector<int>& nums) {
vector<int> result;
for (int i = 0; i < nums.size(); i++) {
// 采用map记录元素出现的次数
if (m.find(nums[i]) != m.end()) {
m[nums[i]] = m[nums[i]] + 1;
}
else {
m[nums[i]] = 1;
}
}
for (map<int, int>::iterator it = m.begin(); it != m.end(); it++) {
if ((*it).second == 1) {
// 将出现一次的元素记录到result里面
result.push_back((*it).first);
}
}
return result;
}
};