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).數組
For example, given [1,2,3,4], return [24,12,8,6].spa
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.)指針
有三種狀況:
數組元素不含0,像[1,2,3,4], return [24,12,8,6]
數組元素有1個0,[1,0,3,4], return [0,12,0,0],是0的那個位置是其餘元素的乘積
數組元素有2個或者2個以上0,[1,0,0,4]則返回[0,0,0,0],返回所有是0.code
public class ProductArrayExceptSelf { public int[] solution(int[] nums) { int zeroCount = 0; for (int n : nums) if (n == 0) zeroCount++; // 有兩個或者兩個以上的元素是0,那麼數組設爲全零返回 if (zeroCount > 1) { for (int i = 0; i < nums.length; i++) nums[i] = 0; } else if (zeroCount == 0) { // 若是沒有0,則計算全部的乘積 int product = 1; for (int n : nums) product *= n; // 每一個數組元素置爲product / 該位置值便可 for (int i = 0; i < nums.length; i++) nums[i] = product / nums[i]; } else { // 若是元素中有1個0 int product = 1; // 跳過那個元素,計算全部的乘積 for (int n : nums) if (n != 0) product *= n; // 元素爲0的位置置爲product,其餘置爲0 for (int i = 0; i < nums.length; i++) if (nums[i] == 0) nums[i] = product; else nums[i] = 0; } return nums; } public static void main(String[] args) { System.out.println(Arrays.toString(new ProductArrayExceptSelf().solution(new int[] { 1, 0, 3, 0 }))); } }
補上one pass 且不用除法, o(n)解法。
使用左右指針,一遍遍歷便可element
public int[] solution2(int[] nums) { int[] result = new int[nums.length]; Arrays.fill(result, 1); int left = 1, right = 1; int len = nums.length; for (int i = 0; i < len; i++) { result[i] *= left; result[len - 1 - i] *= right; left *= nums[i]; right *= nums[len - i - 1]; //System.out.println(Arrays.toString(result)); } return result; }