標籤:lin public pre nal interview you visio ... pts
Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
Example:
Input: [1,2,3,4]Output: [24,12,8,6]
Note: Please solve it without division and in O(n).
Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)
分析:題目翻譯一下: 要求計算一個數組中,除了自己本身之外的所有元素的乘積,不能使用除法。光看要求,不看限制,我們有兩個思路:1、計算所有元素的乘積product,然後從頭到尾遍曆,對於任意的i,res[i] = product / nums[i]。 這當然是最一般的方法,但是不能用除法,這個方法就忽略了。2、加入我們計算i位置的結果res[i],實際就是計算res[i]=nums[0]*nums[1]*...*nums[i-1]*nums[i+1]*...*nums[nums.length-1] 下面就是玄學: res[i]=nums[0]*nums[1]*...*nums[i-1]*nums[i+1]*...*nums[nums.length-1]=(nums[0]*nums[1]*...*nums[i-1])*(nums[i+1]*...*nums[nums.length-1]) 也就是將結果分成兩個部分,left和right,所以res[i]=left[i]*right[i],下面就是找如何確定left和right數組。 left[i]=nums[0]*nums[1]*...*nums[i-1],這就是一個遞迴嘛。,left[i]=left[i-1]*nums[i-1]代碼如下:
1 class Solution { 2 public int[] productExceptSelf(int[] nums) { 3 int n = nums.length; 4 5 int[] left = new int[n]; 6 left[0] = 1; 7 for ( int i = 1 ; i < n ; i ++ ) 8 left[i] = left[i-1] * nums[i-1]; 9 10 int[] right = new int[n];11 right[n-1] = 1;12 for ( int i = n-2 ; i >= 0 ; i -- )13 right[i] = right[i+1] * nums[i+1];14 15 int[] res = new int[n];16 for ( int i = 0 ; i < n ; i ++ )17 res[i] = left[i] * right[i];18 19 return res;20 }21 }
已耗用時間1ms,擊敗100%。
但是這個方法還是用到了多餘的空間,題目說希望可以不使用多餘的空間。因此可以用一個right變數來代替從右向左的那次遍曆。
參考discuss大神的代碼:
1 public int[] productExceptSelf(int[] nums) { 2 int n = nums.length; 3 int[] res = new int[n]; 4 res[0] = 1; 5 for (int i = 1; i < n; i++) { 6 res[i] = res[i - 1] * nums[i - 1]; 7 } 8 int right = 1; 9 for (int i = n - 1; i >= 0; i--) {10 res[i] *= right;11 right *= nums[i];12 }13 return res;14 }
[leetcode] Product of Array Except Self