[LeetCode] 4Sum

[Problem]
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)

[Analysis]

对数组进行排序，使用4个指针，先固定前面前面两个，然后对后面两个使用两边夹逼的方法。例如，对于上面的例子，排序后S = {-2 -1
0 0 1 2}，先固定-2
-1，然后后面连个指针指向0和2，此时的sum=(-2)+(-1)+(0)+(2)=-1，比target小，则第3个指针+1，此时的sum=(-2)+(-1)+(0)+(2)=-1，sum<target，继续移动第3个指针，sum=(-2)+(-1)+(1)+(2)=0，sum==target，保存结果，第3个指针和第4个指针互相靠拢。

此过程中注意剪枝，移动指针的时候，如果移动后的指针指向的数和移动前指向的数相同，则可以直接跳过。

[Solution]

class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int i = 0, j = 0, k = 0, l = 0, m = 0;
vector<vector<int> > res;
if(num.size() < 4)
return res;

// not null
sort(num.begin(), num.end());
while(i <= num.size() - 4){
// the first number is bigger than 0
//if(num[i] > target)break;

// add the second number
j = i + 1;
while(j <= num.size() - 3){
// the sum of the first two number is bigger than 0
//if(num[i] + num[j] > target)break;

// add the third number
k = j + 1;
l = num.size() - 1;
while(l > k){
// the sum of the four number equals to 0, add to result
int sum = num[i] + num[j] + num[k] + num[l];
if(sum == target){
vector<int> tmp;
tmp.push_back(num[i]);
tmp.push_back(num[j]);
tmp.push_back(num[k]);
tmp.push_back(num[l]);
res.push_back(tmp);

// move forward for the same num[l]
m = num[l];
l--;
while(l > k && num[l] == m){
l--;
}

// move forward for the same num[k]
m = num[k];
k++;
while(k < l && num[k] == m){

9784
k++;
}
}
else if(sum > target){
// move forward for the same num[l]
m = num[l];
l--;
while(l > k && num[l] == m){
l--;
}
}
else{
// move forward for same num[k]
m = num[k];
k++;
while(k < l && num[k] == m){
k++;
}
}
}

// move forward for same num[j]
m = num[j];
j++;
while(j <= num.size() - 3 && num[j] == m){
j++;
}
}

// move forward for same num[i]
m = num[i];
i++;
while(i <= num.size() - 4 && num[i] == m){
i++;
}
}
return res;
}
};

说明：版权所有，转载请注明出处。Coder007的博客