(leetcode) 238. Product of Array Except Self My Submissions QuestionEditorial Solution

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:如果这个数组中有大于等于2个0的时候，肯定都是0，如果这个数组中只有一个0，那么除这个为0的位置外，别的位置都为0，对于没有0的情况，先把所以数存起来，当前位置的值，等于乗起来的值除以当前的值。

代码如下(已通过leetcode)

public class Solution {

public int[] productExceptSelf(int[] nums) {

int count0=0;

for(int i=0;i<nums.length;i++)

if(nums[i]==0) count0++;

if(count0>=2) {

for(int i=0;i<nums.length;i++)

nums[i]=0;

} else {

if(count0==1) {

int tempmul=1;

for(int i=0;i<nums.length;i++)

if(nums[i]!=0) tempmul=tempmul*nums[i];

for(int i=0;i<nums.length;i++)

if(nums[i]==0) nums[i]=tempmul;

else nums[i]=0;

} else {

int tempmul=1;

for(int i=0;i<nums.length;i++)

tempmul=tempmul*nums[i];

for(int i=0;i<nums.length;i++)

nums[i]=tempmul/nums[i];

}

}

return nums;

}

}