Leetcode 238. Product of Array Except Self

238. Product of Array Except Self                  

Total Accepted: 51070           Total Submissions: 116543           Difficulty: Medium        

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.)

 

思路：product存储的是所有不为0的数的积。注意有0、1、2个0的情况。

(1) 有0个0：v[i]=product/nums[i];

(2) 有1个0：nums[i]==0,v[i]=product;nums[i]!=0,v[i]=0;

(3) 有2个0：任意i,v[i]=0

 

代码：

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        long long product=1;
        int i,num=0,pos=-1;
        for(i=0;i<nums.size();i++){
            if(nums[i]==0){
               num++;
               pos=i;
               continue;
            }
            product=product*nums[i];
        }
        vector<int> v(nums.size(),0);
        if(num==1){
            v[pos]=product;
        }
        if(num==0){
            for(i=0;i<nums.size();i++){
                v[i]=product/nums[i];
            }
        }
        return v;
    }
};