从数组中选出n个数之和为k

LeetCode15. 3Sum

题目描述：

https://leetcode.com/problems/3sum/

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)

LeetCode15. 3Sum 相当于1. Two Sum的一个变形嘛，题目大意就是从无序数组中找3个数a,b,c，使得a+b+c=0。返回数组中所有的可能作为结果集，结果集中不能包含重复。

思路：

我们可以写成a+b=-c的形式，-c作为target，只不过这次target不是传参指定，而是从给定的数组中找。

遍历i=0到i=len-2，使得target=-nums[i]，然后再用双指针的方法选数,选数的范围[i+1,len-1]。

唯一需要注意的就是去重。

代码：

class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
int len=nums.size();
vector<vector<int>> res;
if(len==0){
return res;
}
sort(nums.begin(),nums.end());
int low,high,target;
for(int i=0;i<len-1;i++){
if(i>0 && nums[i-1]==nums[i]){
continue;
}
target=0-nums[i];
low=i+1;
high=len-1;
while(low<high){
if(nums[low]+nums[high]==target){
vector<int> v;
v.push_back(nums[i]);
v.push_back(nums[low]);
v.push_back(nums[high]);
res.push_back(v);
low++;
high--;
while(low<high && nums[low-1]==nums[low]){
low++;
}
while(low<high && nums[high+1]==nums[high]){
high--;
}
}
else if(nums[low]+nums[high]<target){
low++;
}
else{
high--;
}
}
}
return res;
}
};

题目2

如果是从数组中选出n个数之和为k呢，返回结果集，结果集不含重复。

思路：

我们就用递归回溯来解决问题了。

需要注意的还是去重。

我用pre来处理重复问题，对于递归进入的同一层函数，所选的数字不能相同，所以用pre来记录下标， 如果相同就continue.

代码：

void combinationSum(vector<vector<int>>& res, vector<int>& candidates, int len, int cur, int sum,int n, int target, vector<int>& v){
if (sum > target || cur > len || v.size()>n){
return;
}
else if (sum == target && v.size()==n){
res.push_back(v);
return;
}
else{
int pre = -1;
for (int i = cur+1; i < len; i++){
if (pre != -1 && candidates[pre] == candidates[i]){
continue;
}
pre = i;
sum += candidates[i];
v.push_back(candidates[i]);
combinationSum(res, candidates, len, i + 1, sum,n, target, v);
v.pop_back();
sum -= candidates[i];
}
}
}
vector<vector<int>> combinationSum(vector<int>& candidates, int n, int target){
int len = candidates.size();
vector<vector<int>> res;
if (len == 0){
return res;
}
vector<int> v;
sort(candidates.begin(),candidates.end());
combinationSum(res, candidates, len, -1, 0,n, target, v);
return res;
}

int main()
{
vector<int> candidates;
candidates.push_back(2);
candidates.push_back(1);
candidates.push_back(0);
candidates.push_back(7);
candidates.push_back(1);
candidates.push_back(0);
candidates.push_back(7);
vector<vector<int>> res = combinationSum(candidates,2,9);
system("pause");
return 0;
}