版权声明：由于一些问题，理论类博客放到了blogger上，希望各位看官莅临指教https://efanbh.blogspot.com/；本文为博主原创文章，转载请注明本文来源 https://blog.csdn.net/wyf826459/article/details/81915085
题目描述：
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]
]算法描述：
前面已经做过好多类似的题目，2sum，3sum等，这个题可以看成是原来题目的加强版，这里在2sum题目的基础上改造，即先排序，然后在二重for循环内求2sum的解。在我们选定两个数（a,b）后，题目就变成了在剩余数据中寻找和为target - a -b的2Sum解。
具体：
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> res;
int n = nums.size();
if(n < 4) return res;
vector<int> oneres(4);
sort(nums.begin(),nums.end());
for(int i = 0; i < n - 3;i ++)
{
//优化，如果出现这种情况说明在当前的i之后已经不会再出现合适的解了
if(nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) break;
//优化，如果出现这种情况，说明当前的i值没有对应的解
if(nums[i] + nums[n - 1] + nums[n - 2] + nums[n - 3] < target) continue;
for(int j = i + 1; j < n - 2; j ++)
{
//2sum
//下面两句，类似于前面的优化，解释同上
if(nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) break;
if(nums[i] + nums[j] + nums[n - 1] + nums[n - 2] < target) continue;
int fsum = nums[i] + nums[j];
int bsum = target - fsum;
for(int k = j + 1, m = n - 1; k < m; )
{
if(nums[k] + nums[m] == bsum)
{
oneres[0] = nums[i], oneres[1] = nums[j], oneres[2] = nums[k], oneres[3] = nums[m];
res.push_back(oneres);
while(k < m && nums[k + 1] == nums[k])k ++;
while(k < m && nums[m - 1] == nums[m])m --;
k ++, m --;
}
else if(nums[k] + nums[m] > bsum) m --;
else k ++;
}
while(j < n - 2 && nums[j + 1] == nums[j])j ++;
}
while(i < n - 3 && nums[i + 1] == nums[i])i ++;
}
return res;
}
};