LeetCode 0238 - Product of Array Except Self

Reck Zhang

发布于 2021-08-11 12:00:20

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发布于 2021-08-11 12:00:20

文章被收录于专栏：Reck Zhang

Product of Array Except Self

Desicription

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.)

Solution

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

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原始发表：2018-09-18，如有侵权请联系 cloudcommunity@tencent.com 删除

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