标签：
238. Product of Array Except Self
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.)
题目分析：
给定一个num数组，n>1，输出一个output数组，且output[i]等于除num[i]之外所有元素的乘积，给出一个满足一下条件的solution:
1、 不使用除法
2、 时间复杂度O(n)
3、 不使用额外的储存空间（输出数组不算）
解
假定
s1[0] = nums[0];
s2[n] = nums[n];
构建以下数组
s1[i] = nums[0]* nums[1] * nums[i];
s2[i] = nums[n]* nums[n-1] *...* nums[i];
则可知
output[i] =s1[i-1] * s2[i+1] =/*s1部分*/ nums[0] * nums[1]*...nums[i-1] * nums[i+1]* nums[i+2]* ... nums[n] /*s2部分*/
Solution1
vector<int> productExceptSelf(vector<int>& nums) { int n = nums.size()-1; vector<int> vS1(n+1),vS2(n+1); vector<int> vRst(n+1); int result_s = 1,result_e = 1; for(int i = 1; i<= n; i++) result_s *= nums[i]; vRst[0] = result_s; for(int i = 0; i< n; i++) result_e *= nums[i]; vRst[n] = result_e; vS1[0] = nums[0]; vS2[n] = nums[n]; for(int i = 1; i<= n; i++) { vS1[i] = vS1[i-1] * nums[i]; //由于vS1[0]已知,从vS1[1]开始计算 vS2[n-i] = vS2[n-i+1] * nums[n-i]; //由于vS2[n]已知,从vS2[n-1]开始计算 } for(int i =1; i< n; i++) { vRst[i] = vS1[i-1] * vS2[i+1]; } return vRst; }
分析两个for循环可知：
1、在第i次循环时，vS1[i-1]是已知的，且vRst[i]的值不会对vS2[i+1]造成影响。
2、所以可将vS1[i-1]用一个int类型变量保存，vS2[i+1]的值则保存为vRst[i+1],以满足题目中不开辟额外空间的要求。
给出以下
Solution2
vector<int> productExceptSelf(vector<int>& nums) { int n = nums.size()-1; vector<int> vRst(n+1); int result_s = 1; int s1 = nums[0]; vRst[n] = nums[n]; for(int i= 1; i<=n; i++) vRst[n-i] = vRst[n-i+1] * nums[n-i]; vRst[0] = vRst[1]; for(int i =1; i<n;i++) { vRst[i] = s1 *vRst[i+1]; s1 *= nums[i]; } vRst[n] = s1; return vRst; }
最后是LeetCode Discuss中大犇 给出的答案，比Solution2更快（虽然3个solution Tn = O(n)）
Solution3
vector<int> productExceptSelf(vector<int>& nums) { int n=nums.size(); int fromBegin=1; int fromLast=1; vector<int> res(n,1); for(int i=0;i<n;i++){ res[i]*=fromBegin; fromBegin*=nums[i]; res[n-1-i]*=fromLast; fromLast*=nums[n-1-i]; } return res; }
LeetCode_238_Product of Array Except Self
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原文地址：http://www.cnblogs.com/HellcNQB/p/5380706.html
