题目描述：
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：
注意陷阱，要考虑完全。这里用count来计数，分别处理不同的情况。
- nums数组全为0；
- nums数组只有一个元素为0；
- nums数组有两个以上元素为0；
- nums数组全不为0。
代码实现：
class Solution {
public int[] productExceptSelf(int[] nums) {
int p = 1;
int count = 0;
int len = nums.length;
int t[] = new int[len];
for(int num:nums){
if(num!=0){
p *= num;
}else{
count++;
}
}
if(count==len){
return nums;
}
for(int i=0;i<len;i++){
if(nums[i]==0){
t[i] = count>1?0:p;
return t;
}
nums[i] = p/nums[i];
}
return nums;
}
}