LeetCode 238. Product of Array Except Self C++

238. Product of Array Except Self

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

Example:

Input:  [1,2,3,4]
Output: [24,12,8,6]
Note: Please solve it without division and in O(n).

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

Approach

1. 题目大意很明显就直接讲思路，因为题目有注明说不能使用除，一开始自己没有想出如何不用除可以解决，思路很简单就是将所有结果相乘，然后到哪个数就除掉哪个数即可，当然前面这些操作是不能有零的，所以需要分类讨论，当有一个零和多个零，以及没有零的情况。当只有一个零的时候，那么除了等于零的地方是非零，其他地方都是零，当多个零的时候，所有都是零，没有零就是刚开始的操作。32ms

Code

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int sum = 1, cnt = 0, N = nums.size();
        for (int i = 0; i < N; i++) {
            if (nums[i])
                sum *= nums[i];
            else
                cnt++;
        }
        if (cnt > 1)
            return vector<int>(N, 0);
        vector<int>ans;
        if (cnt == 1) {
            for (int i = 0; i < N; i++) {
                if (nums[i] == 0)ans.push_back(sum);
                else ans.push_back(0);
            }
            return ans;
        }
        for (int i = 0; i < N; i++) {
            ans.push_back(sum / nums[i]);
        }
        return ans;
    }
};

Again

1. 看了一下人家提交的代码，发现可以先前缀乘，然后再后缀乘，那么就可以达到不用除的目的，思路很巧。

Code

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int N = nums.size();
        vector<int> res(N, 1);
        for (int i = 1; i < N; i++) {
            res[i] = res[i - 1] * nums[i - 1];
        }
        int right = 1;
        for (int i = N - 1; i >= 0; i--) {
            res[i] = res[i] * right;
            right =right* nums[i];
        }
        return res;
    }
};