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238. Product of Array Except Self

时间:2016-04-13 11:19:49      阅读:154      评论:0      收藏:0      [点我收藏+]

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238. Product of Array Except Self

 
 
Total Accepted: 41565 Total Submissions: 97898 Difficulty: Medium

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.)

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Code:


int* productExceptSelf(int* nums, int numsSize, int* returnSize) {
    int i,tmp = 1;
    int *arr = (int *)malloc(sizeof(int)*(numsSize));
    int *re = (int *)malloc(sizeof(int)*(numsSize));
 
    arr[numsSize-1] = 1;
    for(i = numsSize-2;i>=0;i--)
        arr[i] = arr[i+1]*nums[i+1];
    for(i = 0;i<numsSize;i++)\
    {
        re[i] = tmp*arr[i];
        tmp*=nums[i];
    }
    *returnSize = numsSize;
    return re;
}

238. Product of Array Except Self

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原文地址:http://www.cnblogs.com/Alex0111/p/5386009.html

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