*LeetCode-Product of Array Except Self
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思想就是两遍 第一遍把这个数左边的乘积乘好 第二遍从右边开始，把这个数右边的乘上来
public class Solution { public int[] productExceptSelf(int[] nums) { int [] res = new int [nums.length]; res [0] = 1; for ( int i = 1; i < nums.length; i ++ ){ res[i] = nums[ i -1 ] * res[ i -1 ]; } int right = 1; for ( int i = nums.length -1; i >= 0; i --) { res[ i ] *= right; right *= nums[ i ]; } return res; }} 0 0
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