力扣（LeetCode） 238.除自身以外数组的乘积（java）_productexceptself

题目

方法一：定义两个新数组

public class ProductExceptSelf {
 
    public int[] productExceptSelf(int[] nums) {
 
        //每个answer[i]的值可以认为是nums[i]左右两侧的乘积
        //定义l和r两个数组分别存储nums[i]左右两侧的乘积
 
        int n = nums.length;
        int[] l = new int[n];
        int[] r = new int[n];
 
        l[0] = 1; //初始位置左侧无值，乘积为1
        for (int i = 1; i < n; i++) {
            l[i] = l[i-1] * nums[i-1];
        }
 
        r[n-1] = 1; //初始位置右侧无值，乘积为1
        for (int i = n-2; i >= 0; i--) {
            r[i] = r[i+1] * nums[i+1];
        }
 
        int[] answer = new int[n];
        for (int i = 0; i < n; i++) {
            answer[i] = l[i] * r[i]; //求结果
        }
 
        return answer;
    }
 
    public static void main(String[] args) {
        ProductExceptSelf self = new ProductExceptSelf();
        //int[] nums = {1,2,3,4};
        int[] nums = {-1,1,0,-3,3};
        System.out.println(Arrays.toString(self.productExceptSelf(nums)));
    }
}

LeetCode测试结果

方法二：对方法一进行空间压缩

public class ProductExceptSelf {
 
    public static void main(String[] args) {
        ProductExceptSelf self = new ProductExceptSelf();
        int[] nums = {1,2,3,4};
        //int[] nums = {-1,1,0,-3,3};
        System.out.println(Arrays.toString(self.productExceptSelf(nums)));
    }
 
    public int[] productExceptSelf(int[] nums) {
 
        //每个answer[i]的值可以认为是nums[i]左右两侧的乘积
        //（ 出于对空间复杂度分析的目的，输出数组不被视为额外空间。）
        //同样的思路，但为了降低空间复杂度，去掉方法1定义的lr两个数组，让answer[i]先充当l数组得到一组值
        //然后在依次得到r的值并直接计算结果赋值到answer[i]
 
        int n = nums.length;
 
        int[] answer = new int[n];
        answer[0] = 1; //初始位置左侧无值，乘积为1
        for (int i = 1; i < n; i++) {
            answer[i] = answer[i-1] * nums[i-1];
        }
 
        int r = 1;
        answer[n-1] = r * answer[n-1];
        for (int i = n-2; i >= 0; i--) {
            r = r * nums[i+1];
            answer[i] = answer[i] * r;
        }
 
        return answer;
    }
 
}

LeetCode测试结果