Product of Array Except Self
Given an array ofnintegers wheren> 1,nums
, return an arrayoutput
such thatoutput[i]
is equal to the product of all the elements ofnums
exceptnums[i]
.
Solve itwithout divisionand 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 arraydoes notcount as extra space for the purpose of space complexity analysis.)
分析
3遍loop, left, right数组叠加乘,不含当前nums[i]。然后ret[i]用left[i]*right[i]。
class Solution {
public int[] productExceptSelf(int[] nums) {
if(nums == null && nums.length == 0){
return null;
}
int n = nums.length;
int[] ret = new int[n];
int[] left = new int[n];
int[] right = new int[n];
for(int i = 0; i < n; i ++){
if(i == 0){
left[i] = 1;
}else{
left[i] = left[i - 1] * nums[i - 1];
}
}
for(int i = n - 1; i >= 0; i --){
if(i == n - 1){
right[i] = 1;
}else{
right[i] = right[i + 1] * nums[i + 1];
}
}
for(int i = 0; i < n; i ++){
ret[i] = left[i] * right[i];
}
return ret;
}
}
constant space complexity,用ret数组直接替代左右数组
class Solution {
public int[] productExceptSelf(int[] nums) {
if(nums == null && nums.length == 0){
return null;
}
int n = nums.length;
int[] ret = new int[n];
int temp = 1;
for(int i = 0; i < n; i ++){
if(i == 0){
ret[i] = 1;
}else{
ret[i] = ret[i - 1] * nums[i - 1];
}
}
for(int i = n - 2; i >= 0; i --){
temp = temp * nuams[i + 1];
ret[i] = ret[i] * temp;
}
return ret;
}
}
Last updated
Was this helpful?