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.)
代码
Approach 1: Left + Right Array
class Solution {
public int[] productExceptSelf(int[] nums) {
int length = nums.length;
int[] L = new int[length];
int[] R = new int[length];
int[] answer = new int[length];
L[0] = 1;
for (int i = 1; i < length; i++) {
L[i] = nums[i - 1] * L[i - 1];
}
R[length - 1] = 1;
for (int i = lenght - 2; i >= 0; i--) {
R[i] = nums[i + 1] * R[i + 1];
}
for (int i = 0; i < length; i++) {
answer[i] = L[i] * R[i];
}
return answer;
}
}
Approach 2: O(1) space approach
class Solution {
public int[] productExceptSelf(int[] nums) {
int length = nums.length;
int[] answer = new int[length];
answer[0] = 1;
for (int i = 1; i < length; i++) {
answer[i] = nums[i - 1] * answer[i - 1];
}
int R = 1;
for (int i = length - 1; i >= 0; i--) {
answer[i] = answer[i] * R;
R *= nums[i];
}
return answer;
}
}