【LeetCode】18. 4Sum 解题小结

题目：

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: 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]
]

这个题目是上一道题目的延伸，再嵌套一次循环确实可以解决这个问题，不过多了两个判断的情况，若最小和大于target，那么停止循环。

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> res;
        int n = nums.size();
        if (nums.size()<4) return res;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size() - 3; i++){
            if (i > 0 && nums[i] == nums[i-1]) continue;
            if (nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target)break;
            if (nums[i]+nums[n-3]+nums[n-2]+nums[n-1] < target)continue;
            for (int j = i+1; j < nums.size() - 2; j++){
                if (j > i+1 && nums[j] == nums[j-1])continue;
                if (nums[i]+nums[j]+nums[j+1]+nums[j+2] > target)break;
                if (nums[i]+nums[j]+nums[n-2]+nums[n-1] < target)continue;
                int left = j + 1;
                int right = n - 1;
                while(left < right){
                    if (nums[i]+nums[j]+nums[left]+nums[right] < target)left++;
                    else if (nums[i]+nums[j]+nums[left]+nums[right] > target)right--;
                    else{
                        res.push_back(vector<int>{nums[i],nums[j],nums[left],nums[right]});
                        do{left++;}while(left<right && nums[left]==nums[left-1]);
                        do{right--;}while(left<right && nums[right]==nums[right+1]);
                    }
                }
            }
        }
        return res;
    }
};

那么问题来了，如果扩展的K个数的和呢？有没有更好的一般性的解决方案呢？在LeetCode上该题的Discussion上有人贴出的代码如下，可以大家参考参考读读：

void KSum(int k, vector<int>& nums, int l, int r, int target, vector<vector<int>>& retVal, vector<int>& cur, int ci )
    {
        int i, mn, mx;
        int km1 = k - 1;

        if ( r-l+1 < k ) return;

        while ( l < r )
        {
            mn = nums[l];
            mx = nums[r];

            // If K minus 1 largest + min < target, move to larger
            if ( ( mn + km1*mx ) < target ) l++;
            // If K minus 1 smaller + max > target, move to smaller
            else if ( ( km1*mn + mx ) > target ) r--;
            // If K * min > target, stop looking
            else if ( k*mn > target ) break;
            // If K * min == target, reached the threshold, check then stop looking
            else if ( k*mn == target )
            {
                if ( ( l + km1 <= r ) && ( mn == ( nums[l+km1] ) ) )
                {
                    for ( i = 0; i < k; i++ ) cur[ci+i] = mn;
                    retVal.push_back( cur );
                }
                break;
            }
            // If K * max < target, stop looking
            else if ( k*mx < target ) break;
            // If K * max == target, reached the threshold, check then stop looking
            else if ( k*mx == target )
            {
                if ( ( l <= r - km1 ) && ( mx == ( nums[r-km1] ) ) )
                {
                    for ( i = 0; i < k; i++ ) cur[ci+i] = mx;
                    retVal.push_back( cur );
                }
                break;
            }
            // If K == 2, we found a match!
            else if ( k == 2 )
            {
                cur[ci] = mn;
                cur[ci+1] = mx;
                retVal.push_back( cur );
                l++;
                while ( ( l < r ) && ( nums[l] == mn ) ) l++;
                r--;
                while ( ( l < r ) && ( nums[r] == mx ) ) r--;
            }
            // Otherwise, convert the problem to a K-1 problem
            else
            {
                cur[ci] = mn;
                KSum( km1, nums, ++l, r, target - mn, retVal, cur, ci+1 );
                while ( ( l < r ) && ( nums[l] == nums[l-1] ) ) l++;
            }
        }
    }