Leetcode:4Sum与3Sum

4Sum:

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)

先对数组排序，之后进行两重for循环，然后对最后的一个数组设两个指针遍历，初始一个指针指向第二重循环后面一个位置，另一个指针则指向数组最后一个元素，因为数组已经排序，根据这四个数的和与target大小比较，来移动这两个指针：如果比目标数值小，则c++，如果比目标数值大则d--。这里注意重复的问题，在一重循环中如果和前面一个数相同则跳过，因为前面的查找肯定包含了本次的情况。

class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {    vector<vector<int> > ret;        if (num.size() == 0) return ret;        sort(num.begin(), num.end());        for (size_t a = 0; a < num.size(); ++a)    {        if (a != 0 && num[a] == num[a - 1])        {            continue;        }                for (size_t b = a + 1; b < num.size(); ++b)        {            if (b != a + 1 && num[b] == num[b - 1])            {                continue;            }                        size_t c = b + 1;            size_t d = num.size() - 1;                        while (c < d)            {                const int sum = num[a] + num[b] + num[c] + num[d];                                if (sum > target)                {                    --d;                }                else if (sum < target)                {                    ++c;                }                else if (c != b + 1 && num[c] == num[c - 1])                {                    ++c;                }                else if (d != num.size() - 1 && num[d] == num[d + 1])                {                    --d;                }                else                {                    vector<int> result;                                        result.push_back(num[a]);                    result.push_back(num[b]);                    result.push_back(num[c]);                    result.push_back(num[d]);                                        ret.push_back(result);                                        ++c;                    --d;                }            }        }    }        return ret;}};

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)

方法基本一样，实现代码如下：

class Solution {public:    vector<vector<int> > threeSum(vector<int> &num) {        vector<vector<int> > result;        if(num.size() <= 2){            return result;        }        sort(num.begin(), num.end());        for(int i = 0; i < num.size()-2; ++i){            if(i > 0 && num[i] == num[i-1]){                continue;            }            int begin = i+1;            int end = num.size()-1;            while(begin < end){                int sum = num[i] + num[begin] + num[end];                if( sum == 0){                    result.push_back({num[i], num[begin], num[end]});                    begin++;                    while(begin < end && num[begin] == num[begin-1]){                        begin++;                    }                    end--;                    while(end > begin && num[end] == num[end+1]){                        end--;                    }                }else if(sum < 0){                    begin++;                }else{                    end--;                }            }        }        return result;    }};