Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
Example:
Input: [1,2,3,4]
Output: [24,12,8,6]
Note: Please solve it without division and in O(n).
Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)
难度: medium
题目:给定一整数组其长度大于1, 返回输出数组其元素由除当前输入元素外所有元素的乘积组成。
思路:先用输出数组从右向左计算累积元素的乘积,然后从左向左计算nums[i] = product(0, i-1) * result(i + 1)
Runtime: 1 ms, faster than 100.00% of Java online submissions for Product of Array Except Self.
Memory Usage: 40.8 MB, less than 84.97% of Java online submissions for Product of Array Except Self.
class Solution {
public int[] productExceptSelf(int[] nums) {
if (null == nums || nums.length < 2) {
return new int[] {1};
}
int n = nums.length;
int[] result = new int[n];
int product = 1;
for (int i = n - 1; i >= 0; i--) {
result[i] = product * nums[i];
product = result[i];
}
result[0] = result[1];
product = nums[0];
for (int i = 1; i < n - 1; i++) {
result[i] = product * result[i + 1];
product *= nums[i];
}
result[n - 1] = product;
return result;
}
}
**粗体** _斜体_ [链接](http://example.com) `代码` - 列表 > 引用
。你还可以使用@
来通知其他用户。