4Sum -- leetcode
2014-12-22 22:11
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Given an array S of n integers, are there elements a,
b, c, and d in S such that a + b +
c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie,
a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.
算法一,在3sum基础上,再套上一层循环。时间复杂度为O(n^3)
算法2: 利用hash查找。时间复杂度O(n^2 log n)。
此算法时间复杂度要优于算法1。表面上看起来,它的运行速度也会快于算法1.
但实则不然,在leetcode上,统计显示,此算法时间已经沦为用java或者python的时间范围内。远不如算法1快。
这可能是因为它存在大量的内存分配请求和内存拷贝的原因,而导致它比算法1慢。
b, c, and d in S such that a + b +
c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie,
a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.
For example, given array S = {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)
算法一,在3sum基础上,再套上一层循环。时间复杂度为O(n^3)
class Solution { public: vector<vector<int> > fourSum(vector<int> &num, int target) { vector<vector<int> > result; if (num.size() < 4) return result; sort(num.begin(), num.end()); for (int i=0; i< num.size(); i++) { if (i && num[i] == num[i-1]) continue; for (int j=i+1; j<num.size(); j++) { if (j>i+1 && num[j] == num[j-1]) continue; int p = j+1, q = num.size()-1; while (p < q) { const int sum = num[i] + num[j] + num[p] + num[q]; if (sum == target) { vector<int> b; b.push_back(num[i]); b.push_back(num[j]); b.push_back(num[p]); b.push_back(num[q]); result.push_back(b); while (++p < q && num[p] == num[p-1]); while (--q > p && num[q] == num[q+1]); } else if (sum < target) ++p; else --q; } } } return result; } };
算法2: 利用hash查找。时间复杂度O(n^2 log n)。
此算法时间复杂度要优于算法1。表面上看起来,它的运行速度也会快于算法1.
但实则不然,在leetcode上,统计显示,此算法时间已经沦为用java或者python的时间范围内。远不如算法1快。
这可能是因为它存在大量的内存分配请求和内存拷贝的原因,而导致它比算法1慢。
class Solution { public: vector<vector<int> > fourSum(vector<int> &num, int target) { sort(num.begin(), num.end()); set<vector<int> > result; unordered_map<int, set<pair<int, int> > > hash; for (int i=0; i<num.size(); i++) { for (int j=i+1; j<num.size(); j++) { if (j>i+1 && num[j] == num[j-1]) continue; const auto iter = hash.find(target-num[i]-num[j]); if (iter != hash.end()) { for (auto &p: iter->second) { vector<int> b = {p.first, p.second, num[i], num[j]}; result.insert(b); } } } for (int j=0; j<i; j++) { if (!j || num[j] != num[j-1]) hash[num[j]+num[i]].insert(make_pair(num[j], num[i])); } } return vector<vector<int> >(result.begin(), result.end()); } };
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