花花酱 LeetCode 18. 4Sum

zxi — Thu, 20 Sep 2018 06:27:38 +0000

Problem

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

Solution 1: Sorting + Binary Search

Time complexity: O(n^3 log n + klogk)

Space complexity: O(k)

C++

// Author: Huahua
class Solution {
public:
  vector> fourSum(vector &num, int target) {
    set> h; 

    sort(num.begin(), num.end());

    int n = num.size();

    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        for(int k = j + 1; k < n; k++) {
          int t = target - num[i] - num[j] - num[k];
          if (t < num[k]) break;
          if (!std::binary_search(num.begin() + k + 1, num.end(), t)) continue;          
          h.insert({num[i], num[j], num[k], t});          
        }
      }
    }
    return vector>(begin(h), end(h));
  }
};

C++ opt

// Author: Huahua
class Solution {
public:
  vector> fourSum(vector &num, int target) {        
    sort(num.begin(), num.end());
    if (target > 0 && target > 4 * num.back()) return {};
    if (target < 0 && target < 4 * num.front()) return {};
    
    set> h;    
    int n = num.size();

    for (int i = 0; i < n; i++) {   
      for (int j = i + 1; j < n; j++) {                
        for(int k = j + 1; k < n; k++) {
          int t = target - num[i] - num[j] - num[k];
          if (t < num[k]) break;
          if (!std::binary_search(num.begin() + k + 1, num.end(), t)) continue;          
          h.insert({num[i], num[j], num[k], t});          
        }           
      }
    }

    return vector>(begin(h), end(h));
  }
};

Solution 2: Sorting + HashTable

Time complexity: O(n^3 + klogk)

Space complexity: O(n + k)

C++

// Author: Huahua
class Solution {
public:
  vector> fourSum(vector &num, int target) {        
    sort(num.begin(), num.end());
    if (target > 0 && target > 4 * num.back()) return {};
    if (target < 0 && target < 4 * num.front()) return {};
    
    unordered_map index;
    for (int i = 0; i < num.size(); ++i)
      index[num[i]] = i;
    
    set> h;    
    int n = num.size();

    for (int i = 0; i < n; i++) {   
      for (int j = i + 1; j < n; j++) {                
        for(int k = j + 1; k < n; k++) {
          int t = target - num[i] - num[j] - num[k];
          if (t < num[k]) break;
          auto it = index.find(t);
          if (it == index.end() || it->second <= k) continue;
          h.insert({num[i], num[j], num[k], t});
        }           
      }
    }

    return vector>(begin(h), end(h));
  }
};

4sum Archives - Huahua's Tech Road

花花酱 LeetCode 18. 4Sum

Problem

Solution 1: Sorting + Binary Search

C++

C++ opt

Solution 2: Sorting + HashTable

C++

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