# 【小白刷leetcode】第15题

动手刷leetcode，正在准备蓝桥，但是本人算法能力一直是硬伤。。。所以做得一直很痛苦。但是不熟练的事情像练吉他一样，就需要慢速，多练。

题目描述

看这个题目，说实在看的不是很懂。索性我们直接来看输入输出。观察输入输出我们可以知道：输入是一个数组，输出是一个list<list。目标其实就是要找到其中的三元组，三元组可以加起来等于零。而且用过的来组成三元组的不能重复使用。也就是每个元素只能使用一次。

暴力解法

首先，我先想到了暴力破解。

每一个元素都访问一遍，用三个for循环。未来防止ijk访问重复，让每一的初始值都为变化的。j=i+1，k=j+1。
这个理论上是行的通的，就是时间超过了限制。时间复杂度为o(n^3)。然后到了一个测试样例过不了。我用idea试了那个测试样例一直在跑，结束不了了。。。

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public static List<List<Integer>> threeSum(int[] nums) {
        //初步看输入输出理解题意，因为我看题目描述压根看不明白
        //输入一个数组，输出三元数组，里面包含的list，每个加起来等于零，而且使用元素不能重复
        int mask =0;
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            for (int j = i+1; j < nums.length; j++) {
                for (int k = j+1; k < nums.length; k++) {
//                    mask = nums[i]+nums[j];
                    if((nums[i]+nums[j]+nums[k])==0){
                        List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[k]));
                        Collections.sort(list);
                        result.add(list);
                    }
                }
            }
        }
        Set<List<Integer>> set = new HashSet<>(result);
        List<List<Integer>> uniqueList = new ArrayList<>(set);
        return uniqueList;
    }

解释一下代码，就是把符合条件的list经过排序放到list中，经过排序是可以后面用来去重。然后添加完了之后，就把list转化为set集合，然后多余的元素都会被去除，然后由集合再转化为list就可以得到最终结果。

测试的mian函数

下面是测试用的main函数。

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    public static void main(String[] args) {
        long start = System.nanoTime();

//        int[] nums = {-1,0,1,2,-1,-4};
//        int[] nums = 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        int[] nums 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//        System.out.println(threeSum(nums));

        System.out.println(threeSum8(nums));
        Util.printCostTime(start);
//        System.out.println(binarySearch(nums,0, nums.length, 2));
    }

第一次优化

然后我就开始尝试优化，肯定要优化掉一层循环,于是我就把最后一层循环用了顺序查找代替，因为一开始数组是无序的，我便想到排序可以优化查询性能，速度是优化了，但是还是过不了很大很多的测试样例。

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public static List<List<Integer>> threeSum2(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            for (int j = i+1; j < nums.length; j++) {
                int find = SequentialSearch(nums,(0-nums[i]-nums[j]));

                if (find!= -1&&find !=i&&find !=j){
                    List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
                    Collections.sort(list);
                    result.add(list);
                }
            }
        }
        Set<List<Integer>> set = new HashSet<>(result);
        List<List<Integer>> uniqueList = new ArrayList<>(set);
        return uniqueList;
    }

public static int SequentialSearch(int[] nums, int target){
        for (int i = 0; i < nums.length; i++) {
            if(nums[i]==target){
                return i;
            }
        }
        return -1;
    }

第二次优化

既然排序了，顺序查询不用排序也可以，而且如果使用二分查找的话更加快得离谱，直接把时间复杂度的一个n转化成logn了。

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public static List<List<Integer>> threeSum4(int[] nums) {

        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            for (int j = nums.length-1; j > i+1; j--) {
                int find = binarySearch(nums,i+1,j,(0-nums[i]-nums[j]));

                if (find!= -1&&find !=i&&find !=j){
                    List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
                    Collections.sort(list);
                    result.add(list);
                }
            }
        }
        Set<List<Integer>> set = new HashSet<>(result);
        List<List<Integer>> uniqueList = new ArrayList<>(set);
        return uniqueList;
    }

  • 效果是过了一些前面过不了的测试样例，但是来到3000个0的时候运行时间超过了4秒钟，无疑又是失败的。
第三次优化

我经过上一次的优化，得到了一个想法：我的for循环块好像已经优化得可以了，3000个0不行，肯定是这种数据太怪了，有没有办法把数据经过清洗然后再传入呢。于是，这两次的优化都是往数据清洗方面靠了。

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public static List<List<Integer>> threeSum1(int[] nums) {
        Set<Integer> nums1 = new HashSet<>();
        for (int num : nums) {
            nums1.add(num);
        }
        Integer[] nums1Array = nums1.toArray(new Integer[0]);

        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums1Array.length; i++) {
            for (int j = i+1; j < nums1Array.length; j++) {
                for (int k = j+1; k < nums1Array.length; k++) {
//                    mask = nums[i]+nums[j];
                    if((nums1Array[i]+nums1Array[j]+nums1Array[k])==0){
                        List<Integer> list = new ArrayList<>(Arrays.asList(nums1Array[i],nums1Array[j],nums1Array[k]));
                        Collections.sort(list);
                        result.add(list);
                    }
                }
            }
        }
        Set<List<Integer>> set = new HashSet<>(result);
        List<List<Integer>> uniqueList = new ArrayList<>(set);
        return uniqueList;
    }

上面的这一个是优化失败的，我想着可以转化成集合去除里面重复的元素，就可以不用算这么多了，但是多余的元素在一定情况下也是有用的，如果你要去除还要进行判断。这个判断就很难进行下去，把数据清洗单独放到前面，确实没有什么好的方法做判断。

既然数据清洗放到上面不行，那我就放到下面的for里面去。但是传统的arrays类型很难进行增删操作，java给了arraylist可以进行方便的增删操作。我们就要先用java8的新特性流来转化数组变成list。然后我再进行判断，如果result中已经含有对应的list，说明这三个元素重复了，就可以进行安全删除。不过我写到一半发现，如果删了的话，for循环的边界就变得比较难以确定了。于是又放弃了。

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public static List<List<Integer>> threeSum3(int[] nums) {
        Arrays.sort(nums);
        List<Integer> list = Arrays.stream(nums)
                .boxed()
                .collect(Collectors.toList());
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < list.size(); i++) {
            for (int j = list.size()-1; j > i+1; j--) {
                int find = Collections.binarySearch(list,(-list.get(i)-list.get(j)));
                if (find!= -1&&find !=i&&find !=j){
                    List<Integer> list1 = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
                    Collections.sort(list1);
                    result.add(list1);
                    if (result.contains(list1)) {
                        list.remove(i);
                        list.remove(j);
                        list.remove(find);
                    }
                }
            }
        }
        Set<List<Integer>> set = new HashSet<>(result);
        List<List<Integer>> uniqueList = new ArrayList<>(set);
        return uniqueList;
    }

这样的数据清洗思路似乎没办法，那我回到把数据清洗放到前面去，用哈希表先存储一遍数组，记录每一个数出现的次数。如果超过某个数那就不把它读到处理的数组中去。

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public static List<List<Integer>> threeSum5(int[] nums) {
        int k = 100; // 允许的最大重复次数
        Map<Integer, Integer> countMap = new HashMap<>();
        for (int num : nums) {
            countMap.put(num, countMap.getOrDefault(num, 0) + 1);
        }
        int index = 0,cnt=0;
        for (int num : nums) {
            if (countMap.get(num) <= k) {
                nums[index++] = num;
                cnt++;
            }
        }

        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            for (int j = nums.length-1; j > i+1; j--) {
                int find = binarySearch(nums,i+1,j,(0-nums[i]-nums[j]));

                if (find!= -1&&find !=i&&find !=j){
                    List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
                    Collections.sort(list);
                    result.add(list);
                }
            }
        }
        Set<List<Integer>> set = new HashSet<>(result);
        List<List<Integer>> uniqueList = new ArrayList<>(set);
        return uniqueList;
    }

这种其实只能防止一些特殊的测试样例，比如全是0的，但是如果别人刚好需要这么多个，你把别人都清洗掉，就会出现不对。

第四次优化

这一次gpt给了我一点思路，我如果要去掉重复元素的话，可以在处理完之后添加跳过语句，这样就不会影响程序的正常判断了。

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public static List<List<Integer>> threeSum6(int[] nums) {

        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i]>0)break;

            for (int j = nums.length-1; j > i+1; j--) {
                int find = binarySearch(nums,i+1,j,(0-nums[i]-nums[j]));

                if (find!= -1&&find !=i&&find !=j){
                    List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
                    Collections.sort(list);
                    result.add(list);
                }
                //必须放到后面不能放到if之前
                while (j>i+3&&nums[j]==nums[j-1])j--;
            }
            //这个也是
            while (i<nums.length-3&&nums[i]==nums[i+1])i++;
        }
        Set<List<Integer>> set = new HashSet<>(result);
        List<List<Integer>> uniqueList = new ArrayList<>(set);
        return uniqueList;

这一个解已经可以accept了，但是姿势十分扭曲，哈哈哈哈哈。

第五次优化

我看了一下别人的题解，用到了双指针、排序、跳过。只用了30ms，解和解的差距怎么这么大？人和人的差距这么大？

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public static List<List<Integer>> threeSum7(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > 0) break;
            if (i > 0 && nums[i] == nums[i - 1]) continue; // 避免重复计算

            int left = i + 1, right = nums.length - 1;
            while (left < right) {
                int sum = nums[left] + nums[right] + nums[i];
                if (sum == 0) {
                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    while (left < right && nums[left] == nums[left + 1]) left++; // 跳过重复元素
                    while (left < right && nums[right] == nums[right - 1]) right--; // 跳过重复元素
                    left++;
                    right--;
                } else if (sum < 0) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return new ArrayList<>(new LinkedHashSet<>(result));
    }

第六次优化

这一个只用了20ms，超过了百分之九十的人，解和解的差距怎么这么大？人和人的差距这么大？

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public static List<List<Integer>> threeSum8(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        for (int i = 0; i < nums.length - 2; i++) {
            if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) { // 避免重复计算
                int left = i + 1, right = nums.length - 1, target = -nums[i];
                while (left < right) {
                    if (nums[left] + nums[right] == target) {
                        result.add(Arrays.asList(nums[i], nums[left], nums[right]));
                        while (left < right && nums[left] == nums[left + 1]) left++; // 跳过重复元素
                        while (left < right && nums[right] == nums[right - 1]) right--; // 跳过重复元素
                        left++;
                        right--;
                    } else if (nums[left] + nums[right] < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }
        return result;
    }
第七次优化

这个是最强的解，没有之一，只是用了3ms,写得我都看不懂。

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public List<List<Integer>> threeSum(int[] nums) {
    
      return nSum(nums, 3, 0);
    }

    public List<List<Integer>> nSum(int[] nums, int k, int target) {
      return new java.util.AbstractList<List<Integer>>() {
        final List<List<Integer>> res = new ArrayList<>();
        final List<Integer> path = new ArrayList<>();
        long min;

        public List<Integer> get(int index) {
          init();
          return res.get(index);
        }

        public int size() {
          init();
          return res.size();
        }

        public void init() {
          if (res.isEmpty()) {
            int n = nums.length;
            long[] Arr = new long[n];
            Arrays.sort(nums);
            min = nums[0];
            for (int i = 0; i < n; i++) {
              Arr[i] = nums[i] - min;
            }
            long NewTarget = (long) target - (long) k * min;
            C(false, Arr, n, k, NewTarget);
          }
        }

        //C(n, k) = C(n - 1, k) + C(n - 1, k - 1)
        public void C(boolean T, long[] a, int n, int k, long target) {
          if (n == 0 || k == 0) {
            if (target == 0 && k == 0) {
              res.add(new ArrayList<>(path));
            }
            return;
          }
          if (k == 2) {
            if (!T && n != a.length && a[n] == a[n - 1]) {
              return;
            }
            //两数之和模板
            twoSum(a, 0, n - 1, target);
            return;
          }
          if (n == k) {
            if (!T && n != a.length && a[n] == a[n - 1]) {
              return;
            }
            //数组中元素和是否等于target
            sumArr(a, n, target);
            return;
          }
          if (check(a, n, k, target)) {
            return;
          }
          C(false, a, n - 1, k, target);
          if (!T && n != a.length && a[n] == a[n - 1]) {
            return;
          }
          if (target - a[n - 1] >= 0) {
            path.add((int) (a[n - 1] + min));
            C(true, a, n - 1, k - 1, target - a[n - 1]);
            path.remove(path.size() - 1);
          }
        }

        void twoSum(long[] a, int l, int r, long target) {
          if (l >= r || a[r - 1] + a[r] < target || a[l] + a[l + 1] > target) {
            return;
          }
          while (r > l) {
            long sum = a[l] + a[r];
            if (sum < target) {
              l++;
            } else if (sum > target) {
              r--;
            } else {
              path.add((int) (a[l] + min));
              path.add((int) (a[r] + min));
              res.add(new ArrayList<>(path));
              path.remove(path.size() - 1);
              path.remove(path.size() - 1);
              while (r > l && a[l] == a[l + 1]) {
                l++;
              }
              while (r > l && a[r] == a[r - 1]) {
                r--;
              }
              l++;
              r--;
            }
          }
        }

        void sumArr(long[] a, int n, long target) {
          for (int i = n - 1; i > -1; i--) {
            target -= a[i];
            path.add((int) (a[i] + min));
          }
          if (target == 0) {
            res.add(new ArrayList<>(path));
          }
          for (int i = n - 1; i > -1; i--) {
            target += a[i];
            path.remove(path.size() - 1);
          }
        }

        boolean check(long[] a, int n, int k, long target) {
          if (n - k < 0) {
            return true;
          }
          long max = 0;
          long min = 0;
          for (int i = 0; i < k; i++) {
            min += a[i];
            max += a[n - i - 1];
          }
          if (target < min || target > max) {
            return true;
          }
          return false;
        }
      };
    }