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;    }};