zoukankan      html  css  js  c++  java
  • [LeetCode] 4Sum

    Well, this problem has a O(n^3) solution similar to 3Sum. That is, fix two elements nums[i] and nums[j] (i < j) and search in the remaining array for two elements that sum to the target - nums[i] - nums[j]. Since i and j both have O(n) possible values and searching in the remaining array for two elements (just like 3Sum that fixes one and search for two other) has O(n) complexity using two pointers (left and right), the total time complexity is O(n^3).

    The code is as follows, which should explain itself.

     1     vector<vector<int> > fourSum(vector<int>& nums, int target) {
     2         sort(nums.begin(), nums.end());
     3         vector<vector<int> > res;
     4         for (int i = 0; i < (int)nums.size() - 3; i++) {
     5             for (int j = i + 1; j < (int)nums.size() - 2; j++) {
     6                 int left = j + 1, right = nums.size() - 1;
     7                 while (left < right) {
     8                     int temp = nums[i] + nums[j] + nums[left] + nums[right];
     9                     if (temp == target) {
    10                         vector<int> sol(4);
    11                         sol[0] = nums[i];
    12                         sol[1] = nums[j];
    13                         sol[2] = nums[left];
    14                         sol[3] = nums[right];
    15                         res.push_back(sol);
    16                         while (left < right && nums[left] == sol[2]) left++;
    17                         while (left < right && nums[right] == sol[3]) right--;
    18                     }
    19                     else if (temp < target) left++;
    20                     else right--;
    21                 }
    22                 while (j + 1 < (int)nums.size() - 2 && nums[j + 1] == nums[j]) j++;
    23             }
    24             while (i + 1 < (int)nums.size() - 3 && nums[i + 1] == nums[i]) i++;
    25         }
    26         return res;
    27     }

    In fact, there is also an O(n^2logn) solution by storing all the possible sum of a pair of elements in nums first. There are such solutions in solution 1, solution 2 and solution 3. You may refer to them. One thing to mention is that all these solutions are slower than the above O(n^3) solution on the OJ :)

    Personally I like solution 3 and rewrite it below.

     1     vector<vector<int> > fourSum(vector<int>& nums, int target) {
     2         sort(nums.begin(), nums.end());
     3         unordered_map<int, vector<pair<int, int> > > mp;
     4         for (int i = 0; i < nums.size(); i++)
     5             for (int j = i + 1; j < nums.size(); j++)
     6                 mp[nums[i] + nums[j]].push_back(make_pair(i, j));
     7         vector<vector<int> > res;
     8         for (int i = 0; i < (int)nums.size() - 3; i++) {
     9             if (i && nums[i] == nums[i - 1]) continue;
    10             for (int j = i + 1; j < (int)nums.size() - 2; j++) {
    11                 if (j > i + 1 && nums[j] == nums[j - 1]) continue;
    12                 int remain = target - nums[i] - nums[j];
    13                 if (mp.find(remain) != mp.end()) {
    14                     for (auto itr = mp[remain].begin(); itr != mp[remain].end(); itr++) {
    15                         int k = (*itr).first, l = (*itr).second;
    16                         if (k > j) {
    17                             vector<int> ans(4);
    18                             ans[0] = nums[i];
    19                             ans[1] = nums[j];
    20                             ans[2] = nums[k];
    21                             ans[3] = nums[l];
    22                             if (res.empty() || ans != res.back())
    23                                 res.push_back(ans);
    24                         }
    25                     }
    26                 }
    27             }
    28         }
    29         return res;
    30     }
  • 相关阅读:
    李永乐,皇帝的新衣背后,共有知识和公共知识
    汇率原理
    mybatis pageHelper 分页插件使用
    oracle中的exists 和not exists 用法详解
    Webservice入门简单实例
    java-可逆加密算法
    idea 卡顿问题
    idea svn操作
    HttpServletrequest 与HttpServletResponse总结
    Spring boot中应用jpa jpa用法
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4567868.html
Copyright © 2011-2022 走看看