238. Product of Array Except Self

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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.)

res数组维护左边的乘积

right变量维护右边的乘积

所以 res[i] = res[i] * right

这样就可以在O(n)时间复杂度、O(1)空间复杂度内计算出结果了

 1 class Solution {
 2 public:
 3     vector<int> productExceptSelf(vector<int>& nums) {
 4         vector<int> res(nums.size(),0);
 5         //保存左边的乘积
 6         res[0] = 1;
 7         for(int i = 1; i < nums.size(); i++){
 8             res[i] = res[i-1] * nums[i-1];
 9         }
10         //计算右边的乘积
11         int right = 1;
12         for(int i = nums.size() - 1; i >= 0 ; i--){
13             res[i] *= right;
14             right *=  nums[i];
15         }
16         return res;
17     }
18 };

238. Product of Array Except Self

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原文地址：http://www.cnblogs.com/hujian1994/p/5141471.html