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.)
public class Solution { public int[] productExceptSelf(int[] nums) { int[] a = new int[nums.length]; int[] b = new int[nums.length]; int[] res = new int[nums.length]; a[0] = 1; for(int i=1;i<nums.length;i++) { a[i] = a[i-1]*nums[i-1]; } b[nums.length-1] = 1; for(int j=nums.length-2;j>=0;j--) { b[j] = b[j+1]*nums[j+1]; } for(int s=0;s<nums.length;s++) { res[s] = a[s]*b[s]; } return res; } }
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.)
public class Solution { public int[] productExceptSelf(int[] nums) { //first pass, {1,1,2,6} //second pass, {24,12,4,1} int[] res = new int[nums.length]; res[0] = 1; for(int i=1; i< nums.length;i++) { res[i] = res[i-1]*nums[i-1]; } int right = 1; for(int i=nums.length-1;i>=0;i--) { res[i] *= right; right *= nums[i]; } return res; } }
reference:https://leetcode.com/discuss/46104/simple-java-solution-in-o-n-without-extra-space
转载于:https://www.cnblogs.com/hygeia/p/5085331.html
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