LeetCode -- Product of Array Except Self

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].


输入数组Nums，返回一个数组output，要求output[i]为，nums[0]...nums[i-1],nums[i+1]....nums[n]的乘积。


思路：
1.区分是否包含0的情况，存0的索引。
2.考虑越界的情况


实现代码：

public class Solution {    public int[] ProductExceptSelf(int[] nums) {    if(nums == null || nums.Length == 1){return nums;}var output = new int[nums.Length];var zeroIndexes = new List<int>();for(var i =0 ;i < nums.Length; i++){    if(nums[i] == 0){        zeroIndexes.Add(i);    }    output[i] = 0;}if(zeroIndexes.Count > 1){    return output;}else if(zeroIndexes.Count == 1){    var s = 1;    for(var i =0 ;i < nums.Length ;i++){        if(i == zeroIndexes[0]){            continue;           }        s *= nums[i];    }       output[zeroIndexes[0]] = s;    return output;}else{    output[0] = 1;    for(var i = 1;i < nums.Length; i++){output[0] *= nums[i];}for(var i = 1; i < nums.Length; i++){    try{output[i] = output[0] * nums [0] % nums[i] == 0 ? output[0] * nums[0] / nums[i] : nums[i] * output[0]/ nums[0];}catch(System.OverflowException ex){output[i] = 0;}}}return output;    }}