【LEETCODE】58、數組分類，適中級別，題目：238、78、287

package y2019.Algorithm.array.medium;

import java.util.Arrays;

/**
 * @ProjectName: cutter-point
 * @Package: y2019.Algorithm.array.medium
 * @ClassName: ProductExceptSelf
 * @Author: xiaof
 * @Description: TODo 238. Product of Array Except Self
 * 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].
 * Input:  [1,2,3,4]
 * Output: [24,12,8,6]
 *
 * 給定長度爲 n 的整數數組 nums，其中 n > 1，返回輸出數組 output ，其中 output[i] 等於 nums 中除 nums[i] 之外其餘各元素的乘積。
 * 來源：力扣（LeetCode）
 * 鏈接：https://leetcode-cn.com/problems/product-of-array-except-self
 * 著作權歸領釦網絡所有。商業轉載請聯繫官方授權，非商業轉載請註明出處。
 *
 * @Date: 2019/7/17 9:31
 * @Version: 1.0
 */
public class ProductExceptSelf {

    public int[] solution(int[] nums) {

        //直接對所有數據求乘積,然後每個數據遍歷的時候，求除數
        int[] res = new int[nums.length];
        int allX = 1, zeroNum = 0;
        for(int i = 0; i < nums.length; ++i) {
            if(nums[i] != 0) {
                allX *= nums[i];
            } else {
                ++zeroNum;
            }
        }

        //求各個位置的值
        if(zeroNum <= 1) {
            for(int i = 0; i < res.length; ++i) {
                if(nums[i] == 0 && zeroNum == 1) {

                    res[i] = allX;
                } else if (zeroNum == 0) {
                    res[i] = allX / nums[i];
                }
            }
        }

        return res;
    }

    public static void main(String[] args) {
        int data[] = {1,0};
        ProductExceptSelf fuc = new ProductExceptSelf();
        System.out.println(fuc.solution(data));
        System.out.println();
    }

}

 

package y2019.Algorithm.array.medium;

import java.io.*;
import java.util.*;

/**
 * @ProjectName: cutter-point
 * @Package: y2019.Algorithm.array.medium
 * @ClassName: Subsets
 * @Author: xiaof
 * @Description: TODO 78. Subsets
 * Given a set of distinct integers, nums, return all possible subsets (the power set).
 * Note: The solution set must not contain duplicate subsets.
 *
 * Input: nums = [1,2,3]
 * Output:
 * [
 *   [3],
 *   [1],
 *   [2],
 *   [1,2,3],
 *   [1,3],
 *   [2,3],
 *   [1,2],
 *   []
 * ]
 *
 *
 *
 * @Date: 2019/7/17 10:49
 * @Version: 1.0
 */
public class Subsets {

    public List<List<Integer>> solution(int[] nums) {
        //輸出所有可能組合，因爲涉及到長度的變化，這裏考慮用遞歸
        List<List<Integer>> res = new ArrayList<>();
        res.add(new ArrayList<>());
        //每次遞歸深度加一，相當於去探索，長度不能超過總長，每次遞歸都是前面一次的集合加上下一次的集合
        //那就要對位置做標記，但是長度又是不固定的並且不重複，那麼考慮用set做標記
        Set mark = new HashSet();
        Arrays.sort(nums);
        //這裏還涉及一個問題，那就是可能有重複的組合，順序不一樣而已，那麼爲了排除掉亂序的，我們對數組拍個順，然後每次只看後面的數據
        allZhuHe(new ArrayList<>(), mark, res, 1, nums, 0);

        return res;
    }

    public void allZhuHe(List<Integer> curList, Set marks, List<List<Integer>> res, int len, int[] nums, int startIndex) {
        if(len > nums.length) {
            return;
        }
        //如果再合理範圍內，那麼我們取不在集合中的數據
//        Set tempSet = new HashSet(marks);
        for(int i = startIndex; i < nums.length; ++i) {
            if(!marks.contains(nums[i])) {
                //如果不包含
                List<Integer> tempList = new ArrayList<>(curList);
                tempList.add(nums[i]);
                res.add(tempList);
                marks.add(nums[i]);
                allZhuHe(tempList, marks, res, len + 1, nums, i);
                marks.remove(nums[i]);
            }
        }
    }

//    public List deepClone(List<Integer> curList) throws IOException, ClassNotFoundException {
//        //直接拷貝對象
//        ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
//        ObjectOutputStream objectOutputStream = new ObjectOutputStream(byteArrayOutputStream);
//        objectOutputStream.writeObject(curList);
//
//        ByteArrayInputStream byteIn = new ByteArrayInputStream(byteArrayOutputStream.toByteArray());
//        ObjectInputStream in = new ObjectInputStream(byteIn);
//        @SuppressWarnings("unchecked")
//        List dest = (List) in.readObject();
//        return dest;
//
//    }

    public static void main(String[] args) {
        int data[] = {1,2,3};
        Subsets fuc = new Subsets();
        System.out.println(fuc.solution(data));
        System.out.println();
    }
}

 

package y2019.Algorithm.array.medium;

/**
 * @ProjectName: cutter-point
 * @Package: y2019.Algorithm.array.medium
 * @ClassName: FindDuplicate
 * @Author: xiaof
 * @Description: TODO 287. Find the Duplicate Number
 * Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive),
 * prove that at least one duplicate number must exist. Assume that there is only one duplicate number,
 * find the duplicate one.
 *
 * Input: [1,3,4,2,2]
 * Output: 2
 *
 * 不能更改原數組（假設數組是隻讀的）。
 * 只能使用額外的 O(1) 的空間。
 * 時間複雜度小於 O(n2) 。
 * 數組中只有一個重複的數字，但它可能不止重複出現一次。
 *
 * @Date: 2019/7/17 11:29
 * @Version: 1.0
 */
public class FindDuplicate {

    public int solution(int[] nums) {
        //明顯就是hash了，但是不能修改原數組，因爲數組再1~n之間，對於這種有範圍的數組，直接hash不用想了
        int[] hashNum = new int[nums.length];
        for(int i = 0; i < nums.length; ++i) {
            if(hashNum[nums[i]] == 0) {
                hashNum[nums[i]]++;
            } else {
                return nums[i];
            }
        }

        return -1;

    }

}