标签：out mission ret 方法 ++ lex ota ack war
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.)
Solution1:
Use two vectors to record the product of numbers befor/after the current number.
The result[i] is the product of fwd[i] * bwd[i]
1 class Solution { 2 public: 3 vector<int> productExceptSelf(vector<int>& nums) { 4 int n = nums.size(); 5 vector<int> fwd(n, 1); 6 vector<int> bwd(n, 1); 7 vector<int> res(n); 8 //forward 9 for(int i = 0; i < n - 1; i++){ 10 fwd[i + 1] = fwd[i] * nums[i]; 11 } 12 //backward 13 for(int i = n - 1; i > 0; i--){ 14 bwd[i - 1] = bwd[i] * nums[i]; 15 } 16 for(int i = 0; i < n; i++){ 17 res[i] = fwd[i] * bwd[i]; 18 } 19 return res; 20 } 21 };
Solution 2:
我们可以对上面的方法进行空间上的优化，由于最终的结果都是要乘到结果res中，所以我们可以不用单独的数组来保存乘积，而是直接累积到res中，我们先从前面遍历一遍，将乘积的累积存入res中，然后从后面开始遍历，用到一个临时变量right，初始化为1，然后每次不断累积，最终得到正确结果
1 class Solution { 2 public: 3 vector<int> productExceptSelf(vector<int>& nums) { 4 vector<int> res(nums.size(), 1); 5 for (int i = 1; i < nums.size(); ++i) { 6 res[i] = res[i - 1] * nums[i - 1]; 7 } 8 int right = 1; 9 for (int i = nums.size() - 1; i >= 0; --i) { 10 res[i] *= right; 11 right *= nums[i]; 12 } 13 return res; 14 } 15 };
238. Product of Array Except Self
标签：out mission ret 方法 ++ lex ota ack war
原文地址：http://www.cnblogs.com/93scarlett/p/6362284.html
