面试经典 150 题 （十四）

就用除法

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] answer = new int[nums.length];
        int sum = 1;
        int zeroNum = 0;
        for (int i = 0; i < nums.length; i++){
            if (nums[i] == 0) zeroNum++;
            sum *= nums[i];
            answer[i] = 0;//初始化
        }
        if (zeroNum >= 2){
            return answer;
        }else if(zeroNum == 1){
            sum = 1;
            for (int i = 0; i < nums.length; i++){
                if (nums[i] == 0){
                    for (int j = 0; j < nums.length; j++){
                        if (j != i){
                            sum *= nums[j];
                        }
                    }
                    answer[i] = sum;
                    break;
                }
            }
        }else{
            for (int i = 0; i < nums.length; i++){
                answer[i] = sum / nums[i];
            }
        }

        return answer;
    }
}

前缀、后缀

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int length = nums.length;
        int[] L = new int[length];  //存放对应元素左侧的乘积
        int[] R = new int[length];  //存放对应元素右侧的乘积
        int num = 1;
        L[0] = 1;
        for (int i = 1; i < length; i++){
            L[i] = L[i - 1] * nums[ i - 1];
        }
        num = 1;
        R[length - 1] = 1;
        for (int j = length - 2; j >= 0; j--){
            R[j] = R[j + 1] * nums[j + 1];
        }
        int[] answer = new int[length];
        for (int i = 0; i < length; i++){
            answer[i] = R[i] * L[i];
        }
        return answer;
    }
}

上述方法的改进，L R 和 answer可以公用一个数组,内存占用可以少个2M

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int length = nums.length;
        int[] L = new int[length];  //存放对应元素左侧的乘积
        //int[] R = new int[length];  //存放对应元素右侧的乘积
        int num = 1;
        L[0] = 1;
        for (int i = 1; i < length; i++){
            L[i] = L[i - 1] * nums[ i - 1];
        }
        num = 1;
        //R[length - 1] = 1;
        int Rtemp = 1;
        for (int j = length - 2; j >= 0; j--){
            //R[j] = R[j + 1] * nums[j + 1];
            Rtemp = Rtemp * nums[j + 1];
            L[j] *= Rtemp;
        }
        // int[] answer = new int[length];
        // for (int i = 0; i < length; i++){
        //     answer[i] = R[i] * L[i];
        // }
        return L;
    }
}