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.)
思路分析:基本思路是用兩個數組left和right,left保存從最左側到當前數之前的所有數字的乘積,right保存從最右側到當前之後的所有數字的乘積,然後,結果數組就是把這兩個數組對應位置相乘即可。如果想只是用O(1)的space,就需要複用輸入和輸出數組的空間,基本思路是,先計算right數組,利用result數組的空間保存,然後計算left數組,只需要一個變量left保存當前從最左側到當前數字之前的所有數字之和。總之,使用好輸入輸出數組的空間,避免使用更多space。以下注釋部分code給出了O(n) 空間複雜度的解法,非註釋部分給出了O(1)空間複雜度的解法。時間複雜度就是O(n)。
AC Code:
public class Solution {
public int[] productExceptSelf(int[] nums) {
/*int len = nums.length;
int [] res = new int[len];
if(len < 2) return res;
int [] left = new int[len];
int [] right = new int[len];
left[0] = 1;
right[len - 1] = 1;
for(int i = len - 1; i > 0; i--){
right[i - 1] = right[i] * nums[i];
}
for(int i = 0; i < len - 1; i++){
left[i + 1] = left[i] * nums[i];
}
for(int i = 0; i < len; i++){
res[i] = left[i] * right[i];
}
return res;*/
int len = nums.length;
int [] res = new int[len];
if(len < 2) return res;
res[len - 1] = 1;
for(int i = len - 1; i > 0; i--){
res[i - 1] = res[i] * nums[i];
}
int left = 1;
for(int i = 0; i < len; i++){
res[i] *= left;
left = left * nums[i];
}
return res;
}
}