(java) 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.)

思路:如果这个数组中2个及以上的0，那都是0.如果只有1个0，除了出现0的位置不是0外别的位置都是0.如果没有0，那么这个位置就是所有数字的乘积除以这个位置上的数

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;

}

}