【LeetCode】238.Product of Array Except Self

题目:

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].

题解:

  有点类似于Trapping Rain Water,从左到右遍历元素,计算该元素左方的乘积,再由右向左,计算右方乘积,同时左右两个积相乘放入result中。

Solution 1 ()

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        vector<int> result(n,1);
        for(int i=1; i<n; ++i) {
            result[i] = result[i-1] * nums[i-1];
        }
        int tmp = 1;
        for(int i=n-2; i>=0; --i) {
            tmp *= nums[i+1];
            result[i] *= tmp;  
        }
        return result;            
    }
};

  比较直观的代码,维护左右两个矩阵。(from here)

Solution 2 ()

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

 

  

posted @ 2017-04-18 19:46  Vincent丶丶  阅读(173)  评论(0编辑  收藏  举报