leetcode 238: Product of Array Except Self

Product of Array Except Self

Total Accepted: 938
Total Submissions: 2400

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.)

[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;
}
}