题目
给定数组和target，从数组中找出4个数组成的所有list，使得每个的和都是target，返回的数组中不能有重复的元素。
分析
略（本文写出nSum的通用方法）
C++代码实现
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class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
vector<vector<int>> res = nSum(nums, 0, 4, target);
return res;
}
vector<vector<int>> nSum(vector<int>& nums, int begin, int count, int target){
vector<vector<int>> res;
if(count == 2){
int i = begin;
int j = nums.size() - 1;
while(i < j){
int x = nums[i], y = nums[j];
if(x + y == target){
vector<int> tmp;
tmp.push_back(nums[i]);
tmp.push_back(nums[j]);
res.push_back(tmp);
while(i < nums.size() && nums[i] == x) i++;
while(j >= begin && nums[j] == y) j--;
}
else if(x + y < target) i++;
else j--;
}
}
else
{
for(int i = begin; i < nums.size(); i ++){
if(i > begin && nums[i] == nums[i-1]) continue;
vector<vector<int>> tmp = nSum(nums, i + 1, count - 1, target - nums[i]);
if(!tmp.empty()){
for(int j = 0; j < tmp.size(); j++){
tmp[j].push_back(nums[i]);
res.push_back(tmp[j]);
}
}
}
}
return res;
}
};
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