leetcode18:4Sum(15-10-6)

一.题目

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)

二.思路

其实和前面3sum思路差不多，先对数组进行排序，然后分别用两个指针来固定两个元素nums[i],nums[j]，再对剩下的两个元素求和，与target-nums[i]-nums[j]进行比较，知道找到相等的为止。

时间复杂度为O(N^3).

三.代码（在3sum基础上延伸过来）

class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
int size=nums.size();
vector<vector<int>> result;
if (size<4) return result;
sort(nums.begin(),nums.end());
for(int i=0;i<size;i++)
{
if(i>0 && nums[i]==nums[i-1]) continue;  //a
int tempTarget1=target-nums[i];
for(int j=i+1;j<size;j++)
{
if(j>i+1 && nums[j]==nums[j-1]) continue;   //b
int tempTarget2=tempTarget1-nums[j];
int front=j+1,rear=size-1;
while(rear>front)
{
if(front>j+1 && nums[front] == nums[front-1])  //c
{
front++;
continue;
}
if(rear<size-1 && nums[rear] == nums[rear+1])     //d
{
rear--;
continue;
}
int sum=nums[rear]+nums[front];
if(sum>tempTarget2) rear--;
else if(sum<tempTarget2) front++;
else
{
vector<int> temp;
temp.push_back(nums[i]);
temp.push_back(nums[j]);
temp.push_back(nums[front]);
temp.push_back(nums[rear]);
result.push_back(temp);
front++;rear--;
}
}
}
}
return result;
}
};

注意几个关键的地方：a,b,c,d这样处理就可以避免重复。