Leetcode 4Sum

先来一个暴力解：

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        vector<vector<int> > res;
        vector<int> r;
        r.resize(4);
        sort(num.begin(), num.end());

        int len = num.size();
        for (int i=0; i<len; i++) {
            if (i!=0 && num[i] == num[i-1]) continue;   // skip dup
            for (int j=i+1; j<len; j++) {
                int psum = num[i] + num[j];
                if (num[j] >= 0 && psum > target) break;
                if (j!=i+1 && num[j] == num[j-1]) continue; // skip dup
                r[0] = num[i], r[1] = num[j];
                
                int ptarget = target - psum;

                int p = j+1;
                int q = len - 1;
                while (p < q) {
                    if (num[p] >= 0 && num[p] > ptarget) break;
                    if (num[q] < 0 && num[q] < ptarget) break;
                    if (p != j+1 && num[p] == num[p-1]) {   // skip dup
                        p++;
                        continue;
                    }
                    if (q != len-1 && num[q] == num[q+1]) { // skip dup
                        q--;
                        continue;
                    }
                    int s = num[p] + num[q];
                    if (s < ptarget) {
                        p++;
                    } else if (s > ptarget) {
                        q--;
                    } else {
                        r[2] = num[p];
                        r[3] = num[q];
                        res.push_back(r);
                        p++, q--;
                    }
                }
            }
        }
        return res;
    }
};

有些快速跳出的语句加了反而更慢。

第二轮：

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)

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        sort(num.begin(), num.end());
        
        vector<vector<int> > res;
        int len = num.size();
        if (len < 4) {
            res;
        }
        
        for (int i=0; i<len; i++) {
            if (i!=0 && num[i] == num[i-1]) continue;
            for (int j=i+1; j<len; j++) {
                if (j!=i+1 && num[j] == num[j-1]) continue; 
                int p = j+1;
                int q = len -1;
                int rest = target - num[i] - num[j];
                
                while (p < q) {
                    int t = num[p] + num[q];
                    if (t == rest) {
                        if (!res.empty() 
                            && res.back()[0] == num[i] 
                            && res.back()[1] == num[j] 
                            && res.back()[2] == num[p]
                            && res.back()[3] == num[q]) {
                        } else {
                            res.push_back(vector<int>({num[i], num[j], num[p], num[q]}));
                        }
                        p++;
                    } else if (t > rest) {
                        q--;
                    } else {
                        p++;
                    }
                }
            }
        }
        return res;
    }
};