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LeetCode-238-Product-of-Array-Except-Self.java
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LeetCode-238-Product-of-Array-Except-Self.java
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/*
http://www.programcreek.com/2014/07/leetcode-product-of-array-except-self-java/
Test Case: [0, 0] -> [0, 0]
1.Forward-Backward traversal
O(N) space
2.Forward-Backward traversal
Space O(N)
*/
public class Solution {
// 1.
// public int[] productExceptSelf(int[] nums) {
// if(nums == null) return null;
// int[] result = new int[nums.length];
// int[] fromLeft = new int[nums.length];
// int[] fromRight = new int[nums.length];
// fromLeft[0] = 1;
// fromRight[nums.length - 1] = 1;
// for(int i = 1; i < nums.length; i++){
// fromLeft[i] = fromLeft[i - 1] * nums[i - 1];
// }
// for(int i = nums.length - 2; i >= 0; i--){
// fromRight[i] = fromRight[i + 1] * nums[i + 1];
// }
// for(int i = 0; i < nums.length; i++){
// result[i] = fromLeft[i] * fromRight[i];
// }
// return result;
// }
// 2.
// public int[] productExceptSelf(int[] nums) {
// if(nums == null) return null;
// int[] result = new int[nums.length];
// result[0] = 1;
// for(int i = 1; i < nums.length; i++){
// result[i] = result[i - 1] * nums[i - 1];
// }
// int right = 1;
// for(int i = nums.length - 1; i >= 0; i--){
// result[i] *= right;
// right *= nums[i]; //be careful
// }
// return result;
// }
// 3.
/*
https://leetcode.com/problems/product-of-array-except-self/discuss/65747/How-from-O(N)-to-O(1)
O(1) Space
*/
public int[] productExceptSelf(int[] nums) {
if(nums == null) return null;
int[] result = new int[nums.length];
int left = 1, right = 1;
Arrays.fill(result, 1);
for (int i = 0; i < nums.length; i++) {
result[i] *= left;
result[nums.length - i - 1] *= right;
left *= nums[i];
right *= nums[nums.length - i - 1];
}
return result;
}
}