LeetCode
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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.)
给出一个数组,使该数组的数满足,i位置是其余位置的数相乘而来。
将数组中所有数相乘,再除以每个位置上的数即可。注意0的个数
class Solution {public: vector<int> productExceptSelf(vector<int>& nums) { long ans = 1; int cnt = 0; for (int i = 0; i < nums.size(); ++i) { if (nums[i] == 0) cnt++; else ans *= nums[i]; } if (cnt >= 2) { for (int i = 0; i < nums.size(); ++i) nums[i] = 0; return nums; } if (cnt == 1) { for (int i = 0; i < nums.size(); ++i) { if (nums[i] != 0) nums[i] = 0; else nums[i] = ans; } return nums; } for (int i = 0; i < nums.size(); ++i) { nums[i] = ans / nums[i]; } return nums; }};
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