leetCode No.238 Product of Array Except Self

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本文介绍了一种解决数组中每个元素等于除自身外其余元素乘积的问题的方法，该方法不使用除法，并且时间复杂度为O(n)。通过两次遍历实现，首次遍历记录左侧元素的乘积，第二次遍历更新右侧元素的乘积。

题目

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.)

标签：Array
相似问题： (H) Trapping Rain Water， (M) Maximum Product Subarray, (H) Paint House II

题意

给定一个大小大于1的整型数组，返回一个数组，数组中每个元素的值为除了元素本身外所有元素的乘积。

解题思路

首先从左到右，每个数等于自己左边数的乘积，每个数再乘以自己右边的数，即为结果。

代码

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