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problems:
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:
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)
Solution:依然采用先排序 然后左右夹逼 时间复杂度O(n3) 同2sum
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> result;
if(nums.size()<4) return result;
sort(nums.begin(),nums.end());
auto last=nums.end();
for(auto a=nums.begin();a<prev(last,3);++a)
for(auto b=next(a);b<prev(last,2);++b)
{
auto c=next(b);
auto d=prev(last);
while(c<d)
{
if(*a+*b+*c+*d<target)
++c;
else if(*a+*b+*c+*d>target)
--d;
else{
result.push_back({*a,*b,*c,*d});
++c;
--d;
}
}
}
sort(result.begin(),result.end());
result.erase(unique(result.begin(),result.end()),result.end());
return result;
}
};
尽可能的降低时间复杂度，我们可以用一个hashmap来缓存两个数的和，这样子时间复杂度可以降低到O(n2)
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原文地址：http://www.cnblogs.com/xiaoying1245970347/p/4553868.html
