标签：des blog io strong for ar art div
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:
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)
枚举两个元素，后面两个采取2sum的策略双指针逼近。
1st （９ ｔｒｉｅｓ）
class Solution
{
public:
vector<vector<int> > fourSum(vector<int> &num, int target)
{
//时间复杂度为n的立方
//算法，先定位前2个，然后后两个利用2 sum的思路
//去重复的思路可能不是很好
vector<vector<int> > re;
vector<int> tmpnum;
if(num.size() < 4)
return re;
sort(num.begin(),num.end());
for(int i = 0;i < num.size() - 3;i++)
{
if(i > 0 && num[i] == num[i - 1])
continue;
for(int j = i + 1;j < num.size() - 2;j++)
{
if(j > i+ 1 && num[j] == num[j - 1])
continue;
tmpnum.clear();
tmpnum.push_back(num[i]);
tmpnum.push_back(num[j]);
int tmptarget = target - num[i] - num[j];
int start = j+1;
int end = num.size() - 1;
while(start < end)
{
if(num[start] + num[end] == tmptarget)
{
if(start > j+1 && end < num.size() - 1 && num[start] == num[start - 1] && num[end] == num[end + 1])
{
start++;
end--;
continue;
}
tmpnum.push_back(num[start]);
tmpnum.push_back(num[end]);
re.push_back(tmpnum);
tmpnum.pop_back();
tmpnum.pop_back();
start++;
end--;
}
else if(num[start] + num[end] > tmptarget)
{
end--;
}
else
{
start++;
}
}
}
}
return re;
}
};
2nd （ 3 tries）
class Solution {
public:
vector<vector<int> > ans;
vector<int> res;
vector<vector<int> > fourSum(vector<int> &num, int target) {
int n = num.size();
sort(num.begin(),num.end());
int _1,_2,_3,_4;
for(_1 = 0;_1 < n-3;_1++) {
if(_1 > 0 && num[_1] == num[_1-1])
continue;
for(_2 = _1 + 1;_2 < n-2;_2++) {
if(_2 > _1 + 1 && num[_2] == num[_2-1])
continue;
_3 = _2 + 1;
_4 = n - 1;
while(_3 < _4) {
if(_3 > _2 + 1 && _4 < n-1 && num[_3] == num[_3-1] && num[_4] == num[_4+1]) {
_3++;
_4--;
continue;
}
if(num[_1] + num[_2] + num[_3] + num[_4] == target) {
res.push_back(num[_1]);
res.push_back(num[_2]);
res.push_back(num[_3]);
res.push_back(num[_4]);
ans.push_back(res);
res.clear();
_3++;
_4--;
}
else if(num[_1] + num[_2] + num[_3] + num[_4] > target) {
_4--;
}
else {
_3++;
}
}
}
}
return ans;
}
};
标签：des blog io strong for ar art div
原文地址：http://www.cnblogs.com/weixliu/p/3924349.html
