leetcode 238: Product of Array Except Self

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Product of Array Except Self

Given an array of n integers where n > 1, nums, return an arrayoutput such that output[i] is equal to the product of all the elements ofnums 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]

维持两个数组, left[] and right[]. 分别记录 第i个元素 左边相加的和left[i] and  右边相加的和right[i].  那么结果res[i]即为  left[i]+right[i]. follow up 要求O(1)空间. 利用返回的结果数组,先存right数组. 再从左边计算left,同时计算结果值, 这样可以不需要额外的空间.

[CODE]

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

leetcode 238: Product of Array Except Self

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原文地址：http://blog.csdn.net/xudli/article/details/46911603