三数之和

题目描述：

给出一个有n个整数的数组S，在S中找到三个整数a, b, c，找到所有使得a + b + c = 0的三元组。

 注意事项

在三元组(a, b, c)，要求a <= b <= c。

结果不能包含重复的三元组。

样例

如S = {-1 0 1 2 -1 -4}, 你需要返回的三元组集合的是：

(-1, 0, 1)

(-1, -1, 2)

 

解法1：三层for循环，时间复杂度O(n^3)

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//解法1: O(n^3) class Solution { public:     /**      * @param numbers : Give an array numbers of n integer      * @return : Find all unique triplets in the array which gives the sum of zero.      */     vector<vector<int> > threeSum(vector<int> &nums) {         // write your code here         vector<vector<int> > ans;         int n = nums.size();         map<vector<int>, int> mp;         vector<int> temp;           for (int i=0; i<n; ++i) {             for (int j=i+1; j<n; ++j) {                 for (int k=j+1; k<n; ++k) {                     if (nums[i]+nums[j]+nums[k] == 0) {                        temp.resize(0);                        temp.push_back(nums[i]);                        temp.push_back(nums[j]);                        temp.push_back(nums[k]);                        temp.resize(3);                        sort(temp.begin(), temp.end());                        if (mp[temp]) {                         continue;                        }else {                         mp[temp]++;                         ans.push_back(temp);                        }                     }                 }             }         }         return ans;     } };

 

解法2：将原数组备份，遍历每一个数val，两层for循环查找是否能找到两数之和为-val。注意：三个数相等时、或任意两数相等时的情况和去重。

时间复杂度O(n^2).

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//解法2：O(n^2)   class Solution { public:     /**      * @param numbers : Give an array numbers of n integer      * @return : Find all unique triplets in the array which gives the sum of zero.      */     vector<vector<int> > threeSum(vector<int> &nums) {         // write your code here         vector<int> ori_num;         map<int, int> cnt;         for (int i=0; i<nums.size(); ++i) {             ori_num.push_back(nums[i]);             cnt[nums[i]]++;         }         vector<vector<int> > ans;         sort(nums.begin(), nums.end());         vector<int> temp;         map<vector<int>, int> vector_mp;           for (int i=0; i<nums.size(); ++i) {             for (int j=i+1; j<nums.size(); ++j) {                 int tot = nums[i] + nums[j];                 auto iter = find(ori_num.begin(), ori_num.end(), 0-tot);                 if (nums[i] == 0-tot) {                     if (cnt[nums[i]] < 2) continue;                 }                 if (nums[j] == 0-tot) {                     if (cnt[nums[j]] < 2) continue;                 }                 if (nums[i] == nums[j] && nums[i] == 0-tot) {                     if (cnt[nums[i]] < 3) continue;                 }                 if (iter != ori_num.end()) {                     temp.resize(0);                     temp = {0-tot, nums[i], nums[j]};                     temp.resize(3);                     sort(temp.begin(), temp.end());                     if (vector_mp[temp] == 0)                         ans.push_back(temp);                         vector_mp[temp]++;                 }             }         }         sort(ans.begin(), ans.end());         return ans;     } };

 

解法3：对原数组从小到大排序，从左到右遍历val，然后二分查找两个数之和是-val的。左边界i+1,右边界n-1，如果当前和=-val，加入答案，否则如果小于，左边界右移，相反右边界左移。可以利用map去重，也可以利用set容器不能包含重复项的特点来去重。时间复杂度O(nlog(n)).

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/// 解法3：对原数组从小到大排序，从左到右遍历，然后找两个数之和是-target的。 class Solution { public:     /**      * @param numbers : Give an array numbers of n integer      * @return : Find all unique triplets in the array which gives the sum of zero.      */     vector<vector<int> > threeSum(vector<int> &nums) {         // write your code here         sort(nums.begin(), nums.end());         set<vector<int> > st;           for (int i=0; i<nums.size(); ++i) {             int val = nums[i];             if (val > 0) break;             int l = i + 1;             int r = nums.size() - 1;             while(l < r) {                 int temp = nums[l] + nums[r];                 if (temp == -val) {                     st.insert({val, nums[l], nums[r]});                     while(l<nums.size()-1 && nums[l+1] == nums[l]) l++;                     while(nums[r-1] == nums[r] && r > 0) r--;                 }else if (temp > -val) r -= 1;                 else if (temp < -val) l += 1;             }         }         vector<vector<int> > ans;         set<vector<int> >::iterator iter;           for (iter=st.begin(); iter != st.end(); ++iter) {             ans.push_back(*iter);         }         return ans;     } };

　

变形：最接近的三数之和

描述：

给一个包含 n 个整数的数组 S, 找到和与给定整数 target 最接近的三元组，返回这三个数的和。

 注意事项

只需要返回三元组之和，无需返回三元组本身

样例

例如 S = [-1, 2, 1, -4] and target = 1. 和最接近 1 的三元组是 -1 + 2 + 1 = 2.

和上题解法3思路相同，固定一个指针，二分查找。

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class Solution { public:     /**      * @param numbers: Give an array numbers of n integer      * @param target: An integer      * @return: return the sum of the three integers, the sum closest target.      */     int threeSumClosest(vector<int> nums, int target) {         // write your code here         sort(nums.begin(), nums.end());         int n = nums.size();         int ans, minn = 99999999;           for (int i=0; i<n-2; ++i) {             int l = i + 1;             int r = n - 1;             while(l < r) {                 if (l < i+1) break;                 if (r > n-1) break;                 int temp = nums[l] + nums[r] + nums[i];                 int tmin = abs(temp - target);                   if (tmin == 0) {                     ans = temp;                     return ans;                 }else if (tmin < minn) {                     minn = tmin;                     ans = temp;                 }                 if (temp < target) {                     l += 1;                 }else r -= 1;             }         }         return ans;     } };