4Sum

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.

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)

题意：求一个数组中所有和为target的四元组。
思路：参考3sum的解法，3sum的复杂度为O(n^2), 在3sum外加一层循环即可解4sum,这样的话复杂度为O(n^3)。此题O(n^3)的复杂度即可通过。存在O(n^2*logN)的复杂度算法，第二遍时再探讨。

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());
int sum;
int i, j, k, l;
for (int i = 0; i < nums.size()-3; i++){
if (i>0 && nums[i] == nums[i - 1])
continue;
for (int j = i + 1; j < nums.size() - 2; j++){
if (j>i + 1 && nums[j] == nums[j - 1])
continue;
k = j + 1;
l = nums.size() - 1;
while (k < l){
sum = nums[i] + nums[j] + nums[k] + nums[l];
if (sum == target){
vector<int> r(4);
r[0] = nums[i];
r[1] = nums[j];
r[2] = nums[k];
r[3] = nums[l];
result.push_back(r);
k++;
while (k < l && nums[k] == nums[k - 1])
k++;
l--;
while (l > k && nums[l] == nums[l + 1])
l--;
}
else if (sum > target){
l--;
while (l > k && nums[l] == nums[l + 1])
l--;
}
else{
k++;
while (k < l && nums[k] == nums[k - 1])
k++;
}
}
}
}
return result;
}
};