238. 除自身以外数组的乘积

解法1

假设使用除法

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int cnt0 = 0, idx0 = 0, lc = 1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0) {
                cnt0++;
                idx0 = i;
            } else {
                lc *= nums[i];
            }
        }
        int[] ans = new int[nums.length];
        if (cnt0 == 1) {
            ans[idx0] = lc;
        } else if (cnt0 < 1) {
            for (int i = 0; i < ans.length; i++) {
                ans[i] = lc / nums[i];
            }
        }
        return ans;
    }
}

解法2

使用额外数组

从左向右乘法，从右向左乘法

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return nums;
        }
        int[] left = new int[n];
        int[] right = new int[n];
        int[] ans = new int[n];
        int base = 1;
        for (int i = 0; i < n; i++) {
            base *= nums[i];
            left[i] = base;
        }
        base = 1;
        for (int i = n - 1; i > -1; i--) {
            base *= nums[i];
            right[i] = base;
        }
        ans[0] = right[1];
        ans[n - 1] = left[n - 2];
        for (int i = 1; i < n - 1; i++) {
            ans[i] = left[i - 1] * right[i + 1];
        }
        return ans;
    }
}

解法3

常数项额外空间

left或者right可以作为最终数组，节约空间。

right动态计算，可以只用一个变量。

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return nums;
        }
        int[] left = new int[n];
        left[0] = 1;
        for (int i = 1; i < n; i++) {
            left[i] = left[i - 1] * nums[i - 1];
        }
        int r = 1;
        for (int i = n - 2; i > -1; i--) {
            r *= nums[i + 1];
            left[i] *= r;
        }
        return left;
    }
}