238. Product of Array Except Self

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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].

tip:不能用除法、时间复杂度要求O(n)（这样遍历数组的方法也不行了）

想了好久都没想到O(n)的方法，最后还是用了除法，在有0的情况下还有另外处理，比较复杂，代码如下，仅供忽略

public class Solution {    public int[] productExceptSelf(int[] nums) {        int[] results = new int[nums.length];        int product = 1;        int zeroPos = 0;        for(int i = 0;i<nums.length;i++) {            product *= nums[i];            if(nums[i] == 0) {                zeroPos = i;            }        }        if(product == 0) {            for(int i = 0;i<results.length;i++) {                results[i] = 0;            }            product = 1;            for(int i = 0;i<nums.length;i++) {                if(i == zeroPos) {                    continue;                }                product *= nums[i];            }            results[zeroPos] = product;        } else {            for(int i = 0;i<nums.length;i++) {                results[i] = product/nums[i];            }        }        return results;    }}

正确解法

思路：result[i]的一种求法是：可以先依次乘i的右边各个元素，再乘i的左边各个元素

如题目中的[1,2,3,4],result[1] = 3 * 4 * 1 = 12

public class Solution {    public int[] productExceptSelf(int[] nums) {        int n = nums.length;        int result[] = new int[n];        result[n-1] = 1;        for(int i = n-2;i>=0;i--) {            result[i] = result[i+1] * nums[i+1];        }        int left = 1;        for(int i = 0;i<n;i++) {            result[i] = result[i] * left;            left = left * nums[i];        }        return result;    }}

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