LeetCode:Product of Array Except Self

redis_v發表於2016-01-14

Product of Array Except Self


題目描述:

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.)


思路:

基本思路:把所有元素相乘,再除以對應位置的元素數字,便得到除了自身元素之外,其他所有元素的乘積。

例外情況:若自身元素為0,則不能作為除數。

解決辦法:記錄所有元素的乘積product,非0元素的乘積pro,0元素個數cnt。

    a:若0元素個數>=2,所求所有乘積都為0;

    b:若自身元素不為0,且0元素個數<2,使用product/自身元素,獲得乘積。

    c:若自身元素為0,且0元素個數<2, pro就是所求的乘積。

程式碼:

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
	int n = nums.size();
	int product = 1;
	int pro0 = 1;
	int cnt0 = 0;
	for(int i = 0; i < n; i ++)
	{
		int num = nums[i];
		if(num != 0)
			product *= num;
		else
			++cnt0;
		pro0 *= num;
	}
	for(int i = 0; i < n; i ++)
	{
		if(cnt0 > 1)
			nums[i] = 0;
		else
		{
			int num = nums[i];
			if(num != 0)
			{
				nums[i] = pro0/nums[i];
			}
			else
			{
				nums[i] = product;
			}
		}	
	}
	return nums;
}
};


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