LeetCode – Product of Array Except Self (Java)
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].
Java Solution 1
public int[] productExceptSelf(int[] nums) { int[] result = new int[nums.length]; int[] t1 = new int[nums.length]; int[] t2 = new int[nums.length]; t1[0]=1; t2[nums.length-1]=1; //scan from left to right for(int i=0; i<nums.length-1; i++){ t1[i+1] = nums[i] * t1[i]; } //scan from right to left for(int i=nums.length-1; i>0; i--){ t2[i-1] = t2[i] * nums[i]; } //multiply for(int i=0; i<nums.length; i++){ result[i] = t1[i] * t2[i]; } return result; } |
Java Solution 2
We can directly put the product values into the final result array. This saves the extra space to store the 2 intermediate arrays in Solution 1.
public int[] productExceptSelf(int[] nums) { int[] result = new int[nums.length]; result[nums.length-1]=1; for(int i=nums.length-2; i>=0; i--){ result[i]=result[i+1]*nums[i+1]; } int left=1; for(int i=0; i<nums.length; i++){ result[i]=result[i]*left; left = left*nums[i]; } return result; } |
<pre><code> String foo = "bar"; </code></pre>
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