18. 4Sum
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: 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]
]
解析
class Solution_18 {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> vecs;
if (nums.size()<4)
{
return vecs;
}
sort(nums.begin(), nums.end());
for (int first = 0; first < nums.size() - 3;first++)
{
if (first>0&&nums[first]==nums[first-1])
{
continue;
}
for (int seconde = first + 1; seconde < nums.size() - 2;seconde++)
{
if (seconde>first+1&&nums[seconde]==nums[seconde-1])
{
continue;
}
int third = seconde + 1;
int fouth = nums.size() - 1;
while (third<fouth)
{
vector<int> temp = {nums[first],nums[seconde],nums[third],nums[fouth]};
int sum = accumulate(temp.begin(), temp.end(), 0);
if (sum==target)
{
vecs.push_back(temp);
third++; fouth--; //细节问题
while (third < fouth&&nums[third] == nums[third - 1])
{
third++;
}
while (third < fouth&&nums[fouth] == nums[fouth + 1])
{
fouth--;
}
// third++; fouth--;放在后面有bug
//// start++; end--放在后面就是和后面的数比较
//while (start < end&&nums[start] == nums[start + 1])
//{
// start++;
//}
//while (start < end&&nums[end] == nums[end - 1])
//{
// end--;
//}
//start++; end--;
////或者用do() while()循环
//do{
// k++;
//} while (k < l && ivec[k] == ivec[k - 1]);
//do{
// l--;
//} while (k < l && ivec[l] == ivec[l + 1]);
}
else if(sum<target)
{
third++;
}
else
{
fouth--;
}
}
}
}
return vecs;
}
};
题目来源
