leetcode

3Sum

 Total Accepted: 64333 Total Submissions: 380677My Submissions

Question Solution 

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)

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {       vector<vector<int>> result;       if(nums.size() < 3) return result;       sort(nums.begin(), nums.end());       const int target = 0;       auto last = nums.end();       for(auto i = nums.begin(); i < last - 2; ++i){           if(i > nums.begin() && *i == *(i - 1)) continue;           auto j = i + 1;           auto k = nums.end() - 1;           while(j < k){               if(*i + *j + *k < target){                   ++j;                   while(*j == *(j - 1) && j < k) ++j;               }               else if(*i + *j + *k > target){                   --k;                   while(*k == *(k + 1) && j < k) --k;               }               else{                   result.push_back({*i, *j, *k});                   ++j;                   --k;                   while(*j == *(j - 1) && j < k)  ++j;               }           }       }       return result;    } };

超时的解法，两次排序了用unique去重

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {       vector<vector<int>> result;       if(nums.size() < 3) return result;       sort(nums.begin(), nums.end());       const int target = 0;       auto last = nums.end();       for(auto i = nums.begin(); i < prev(last, 2); ++i){           auto j = next(i);           auto k = prev(last);           while(j < k){               if(*i + *j + *k < target){                   ++j;               }               else if(*i + *j + *k > target){                   --k;               }               else{                   result.push_back({*i, *j, *k});                   ++j;                   --k;               }           }       }       sort(result.begin(), result.end());       result.erase(unique(result.begin(), result.end()), result.end());       return result;    } };

3Sum Closest

 Total Accepted: 44553 Total Submissions: 165635My Submissions

Question Solution 

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).

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        sort(nums.begin(), nums.end());        int result;        bool flag = false;        int min_gap = abs(nums[0] + nums[1] + nums[2] - target);        for(auto i = nums.begin(); i < nums.end() - 2; ++i){            auto j = next(i);            auto k = prev(nums.end());            while(j < k){                int sum = *i + *j + *k;                int gap = abs(sum - target);                if(gap <= min_gap){                    result = sum;                    min_gap = gap;                    if(gap == 0){                        flag = true;                        break;                    }                }                if(sum < target) ++j;                else --k;            }            if(flag)                break;        }        return result;    }};

4Sum
 Total Accepted: 40765 Total Submissions: 188176My Submissions
Question Solution 
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>& nums, int target) {        vector<vector<int>> result;        if(nums.size() < 4) return result;        sort(nums.begin(), nums.end());        for(auto i = nums.begin(); i < nums.end() - 3; ++i){            for(auto j = next(i); j < nums.end() - 2; ++j){                auto k = next(j);                auto l = prev(nums.end());                while(k < l){                    int sum = *i + *j + *k + *l;                    if(sum == target){                        result.push_back({*i, *j, *k, *l});                        do{                            ++k;                        }while(*k == *(k - 1) && k < l);                        do{                            --l;                        }while(*l == *(l + 1) && k < l);                    }                    else if(sum < target)                        ++k;                    else                        --l;                }            }        }        sort(result.begin(), result.end());        result.erase(unique(result.begin(), result.end()), result.end());        return result;    }};