LeetCode —— 3SUM

3Sum

链接：http://leetcode.com/onlinejudge#question_15

原题：

Given an array S of n integers,
are there elements a, b, c in S such
that a + b + c =
0? Find all unique triplets in the array which gives the sum of zero.

Note:

Elements in a triplet (a,b,c)
must be in non-descending order. (ie, a ? b ? c)

The solution set must not contain duplicate triplets.

For example, given array S = {-1 0 1 2 -1 -4},

A solution set is:
(-1, 0, 1)
(-1, -1, 2)

思路：DFS + 剪枝

暴力搜索，复杂度是O（n^3），可以加剪枝

先对数组排序

1）如果 sum + array[i] > 0 那么就不用递归下去了

2）如果 i > start, 并且 array[i] == array[i-1]，就不用递归了，这样还可以避免重复到组合进入最终结果里

做完之后，搜了一下其他人到做法，发现有O(n^2)到算法，还是蛮巧妙到，大概到思路是

先排序，然后对第一个数进行0到size-3遍历，后两个数是两头往中间走。具体可以参考：

http://www.cnblogs.com/etcow/articles/2547108.html

代码：

class Solution {
public:
vector<vector<int> > threeSum(vector<int> &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> vec;
vector<vector<int> > collection;
sort(num.begin(), num.end(), less<int>() );
search(0, 0, 3, num, vec, collection);

return collection;
}

private:
void search(int start, int sum, int count,
const vector<int> &array, vector<int> &vec, vector<vector<int> > &collection) {

if (count == 0) {
if (sum == 0)
collection.push_back(vec);
return;
}

int size = array.size() - count;
for (int i=start; i<=size; i++) {
if (i > start && array[i-1] == array[i])
continue;
if (sum + array[i] > 0)
return;
vec.push_back(array[i]);
search(i+1, sum+array[i], count-1, array, vec, collection);
vec.pop_back();
}
}
};

3Sum Closest

链接：http://leetcode.com/onlinejudge#question_16

原题：

Given an array S of n integers,
find three integers in S such
that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

思路：如果将三个数的和减去target，那么就是求最接近0到组合了，就可以归结为3sum的问题了。

这回尝试用遍历第一个数，然后后两个数从两端往中间走的策略。复杂度为O(n^2)，当然先排序。

代码：

class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (num.size() <= 3) {
int sum = 0;
for (int i=0; i<num.size(); i++)
sum += num[i];
return sum;
}

int min = num[0] + num[1] + num[2] - target;
sort(num.begin(), num.end(), less<int>() );
for (int i=0; i<num.size()-2; i++) {
if (i>0 && num[i] == num[i-1])
continue;
int part = num[i] - target;
for (int m=i+1, n=num.size()-1; m<n; ) {
int sum = part + num[m] + num
;
if (abs(sum) < abs(min))
min = sum;
if (sum > 0)
n--;
else if (sum < 0)
m++;
else
return target;
}
}

return target + min;
}
};

4Sum

链接：http://leetcode.com/onlinejudge#question_18

原题：

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)

思路：直接用 DFS + 剪枝会超时，所以采用一二两个数遍历，三四两个数两头向中间靠拢，

要小心重复组合的加入。

代码:

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
vector<int> vec;
vector<vector<int> > collection;
if (num.size() < 4)
return collection;
sort(num.begin(), num.end(), less<int>() );
search(0, 0, 4, target, num, vec, collection);

return collection;
}

private:
void search(int start, int sum, int count, int target,
const vector<int> &array, vector<int> &vec, vector<vector<int> > &collection) {

if (count == 2) {
for (int m=start, n=array.size()-1; m<n;) {
int temp = sum + array[m] + array
;
if (temp < target)
m++;
else if (temp > target)
n--;
else {
vec.push_back(array[m]);
vec.push_back(array
);
collection.push_back(vec);
vec.pop_back();
vec.pop_back();
m++; n--;
while (m < array.size() && array[m] == array[m-1]) m++;
while (n >= 0 && array
== array[n+1]) n--;
}
}
return;
}

int size = array.size() - count;
for (int i=start; i<=size; i++) {
if (i > start && array[i-1] == array[i])
continue;
vec.push_back(array[i]);
search(i+1, sum+array[i], count-1, target, array, vec, collection);
vec.pop_back();
}
}
};