leetcode--sum集合：2sum，3sum，4sum

k sum：在数组nums中找到k个数相加的和为target，不可重复

这个问题有两个方法：

① double pointer，可以得到O(n^(k-1))

②hash table，得要具体分析一下

2sum：

方法一：

struct num{<span style="white-space:pre"></span>int number;<span style="white-space:pre"></span>int index;};int* twoSum(int* nums, int numsSize, int target) {int *ret = (int*)malloc(2 * sizeof(int));ret[0] = ret[1] = 0;int j;struct num *map = (struct num*)malloc(numsSize*sizeof(struct num));for (j = 0; j<numsSize; j++){map[j].index = j;map[j].number = nums[j];}for (int increment = numsSize / 2; increment > 0; increment /= 2){for (int i = increment; i < numsSize; i++){struct num tmp = map[i];for (j = i; j >= increment; j -= increment){if (tmp.number < map[j - increment].number)map[j] = map[j - increment];elsebreak;}map[j] = tmp;}}int fir = 0, sec = numsSize - 1;while (fir < sec){if (map[fir].number + map[sec].number == target){if (map[fir].index < map[sec].index){ret[0] = map[fir].index + 1;ret[1] = map[sec].index + 1;}else{ret[0] = map[sec].index + 1;ret[1] = map[fir].index + 1;}break;}else if (map[fir].number + map[sec].number < target){fir++;}else{sec--;}}free(map);return ret;}

双指针用while可以写的比较好看，因为要返回的是index，排序之后打乱了，设计一个结构体。

方法二：

int* twoSumI(int* nums, int numsSize, int target) {int *ret = (int*)malloc(2 * sizeof(int));ret[0] = ret[1] = 1;int i = 0, max = 0;for (; i < numsSize; i++){if (nums[i]>max)max = nums[i];}char *hash = (char*)calloc(2*max+1,sizeof(char));for (i = 0; i < numsSize; i++){hash[nums[i]+max] ++;}for (i = 0; i < numsSize; i++){if (hash[target - nums[i] + max] ){if (nums[i] == target / 2 && (hash[target - nums[i] + max]-1)){ret[0] = i + 1;break;}else if (nums[i] == target / 2){continue;}ret[0] = i + 1;break;}}i++;//break 使得最后一次i没有自加for (; i < numsSize; i++){if (nums[i] == target - nums[ret[0] - 1]){ret[1] = i + 1;return ret;}}}

C语言实现的一个简单的hash table，因为题目中说有且只有一个解。可以边记录hash边查找，不过还是要遍历找到index，用C++的话比较方便。

方法三：

class Solution {public:vector<int> twoSum(vector<int>& nums, int target) {map<int, int> mapping;vector<int> ret;int i = 0;while (i<nums.size()){if (mapping.find(target - nums[i]) != mapping.end()){ret.push_back(mapping[target - nums[i]]);ret.push_back(i);return ret;}mapping[nums[i]] = i;i++;}}};

3sum：

方法一：

int** threeSum(int* nums, int numsSize, int* returnSize) {*returnSize = 0;if (numsSize<3)return NULL;int **ret = (int**)malloc(2*numsSize*sizeof(int*));//申请太小，导致run timeint j;for (int increment = numsSize / 2; increment > 0; increment /= 2){for (int i = increment; i < numsSize; i++){int tmp = nums[i];for (j = i; j >= increment; j -= increment){if (tmp < nums[j - increment])nums[j] = nums[j - increment];elsebreak;}nums[j] = tmp;}}for (int i = 0; i < numsSize - 2 ; i++){for (int j = i+1; j < numsSize - 1 && nums[i] + nums[j] <= 0; j++){for (int k = numsSize - 1; k > j && nums[i] + nums[j] + nums[k] >= 0; k--){if (nums[i] + nums[j] + nums[k] == 0){int *col = (int *)malloc(3 * sizeof(int));col[0] = nums[i], col[1] = nums[j], col[2] = nums[k];ret[*returnSize] = col;(*returnSize)++;}while (k - 1>j && nums[k - 1] == nums[k]){k--;}}while (j + 1 < numsSize && nums[j + 1] == nums[j]){j++;}}while (i + 1 < numsSize && nums[i + 1] == nums[i]){i++;}}return ret;}

这是以前写的，也是双指针，写的比较丑，但是逻辑还是挺清晰的~吧

threeSumClosest：

就是要增加一个最接近的判断

int threeSumClosest(int* nums, int numsSize, int target) {int min = INT_MAX;int sum = 0;int ret = 0;int j;for (int increment = numsSize / 2; increment > 0; increment /= 2){for (int i = increment; i < numsSize; i++){int tmp = nums[i];for (j = i; j >= increment; j -= increment){if (tmp < nums[j - increment])nums[j] = nums[j - increment];elsebreak;}nums[j] = tmp;}}for (int i = 0; i < numsSize - 2; i++){int j = i + 1, k = numsSize - 1;while (j < k){sum = nums[i] + nums[j] + nums[k];if (sum == target)return sum;else if (sum < target){j++;}else{k--;}if (target>sum && min>target - sum){min = target - sum;ret = sum;}else if (target<sum && min>sum - target){min = sum - target;ret = sum;}}while (i + 1 < numsSize && nums[i + 1] == nums[i]){i++;}}return ret;}

4sum:

方法一：

int** fourSum(int* nums, int numsSize, int target, int* returnSize) {*returnSize = 0;int **ret = (int **)malloc(numsSize * numsSize*sizeof(int*));int sum = 0;int j;for (int increment = numsSize / 2; increment > 0; increment /= 2){for (int i = increment; i < numsSize; i++){int tmp = nums[i];for (j = i; j >= increment; j -= increment){if (tmp < nums[j - increment])nums[j] = nums[j - increment];elsebreak;}nums[j] = tmp;}}//i,l,j,kfor (int i = 0; i < numsSize - 3; i++){for (int l = i + 1; l < numsSize - 2; l++){int j = l + 1, k = numsSize - 1;while (j < k){sum = nums[i] + nums[l] + nums[j] + nums[k];if (sum == target){int *col = (int *)malloc(4 * sizeof(int));col[0] = nums[i], col[1] = nums[l], col[2] = nums[j],col[3]=nums[k];ret[*returnSize] = col;(*returnSize)++;while (j + 1 < numsSize && nums[j + 1] == nums[j]){j++;}while (k - 1 > j && nums[k - 1] == nums[k]){k--;}j++;k--;}else if (sum < target){j++;}else{k--;}}while (l + 1 < numsSize && nums[l + 1] == nums[l]){l++;}}while (i + 1 < numsSize && nums[i + 1] == nums[i]){i++;}}return ret;}

方法二：

class SolutionI {public:vector<vector<int>> fourSum(vector<int>& nums, int target) {int i, j;unordered_map<int, vector<pair<int, int>>> map;vector<vector<int>> ret;if (nums.size() < 4)return ret;for (i = 0; i < nums.size(); i++){for (j = i+1; j < nums.size(); j++){pair<int, int> index(i, j);map[nums[i] + nums[j]].push_back(index);}}for (auto i = map.begin(); i != map.end();i++){if (map.find(target - i->first) != map.end()){for (auto j = (i->second).begin(); j != (i->second).end(); j++){for (auto k = map[target - i->first].begin(); k != map[target - i->first].end(); k++){if (j->first != k->first &&j->first != k->second &&j->second != k->first &&j->second != k->second){vector<int> four = { nums[(j->first)], nums[j->second], nums[k->first], nums[k->second] };sort(four.begin(), four.end());ret.push_back(four);}}if (i->first == target - i->first){break;}}}}sort(ret.begin(), ret.end());auto tmp = unique(ret.begin(), ret.end());ret.erase(tmp,ret.end());return ret;}};

把四个数分成两个数，将复杂度降为O(n^2)，但是AC的时间是240s，有没有知道 c++ AC中快的是如何实现的呀？？？有个网友也是类似这样的，map记录和检查是否存在放在一起，可是就是超时，不知道为什么，把他的程序也粘出来：

转自：https://leetcode.com/discuss/26678/please-help-why-it-is-time-limit-exceeded-on-this-4sum-problem?show=26678#q26678  

TLE

class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {        vector<vector<int>> rt;        set<vector<int>> mset;        int n = num.size();        if (n<4) return rt;        unordered_map<int, vector<vector<int>> > hm;        for (int i = 0; i<n-1; i++) {            for (int j=i+1; j<n; j++) {                vector<int> v;                v.push_back(i);                v.push_back(j);                unordered_map<int, vector<vector<int>>>::iterator mate = hm.find(target- (num[i]+num[j]));                 if (mate!=hm.end()) {                    for ( vector<vector<int>>::iterator it2 = (mate->second).begin(); it2 != (mate->second).end(); it2++) {                        if ( i!=(*it2)[0] && j!=(*it2)[0] && i!=(*it2)[1] && j!=(*it2)[1] ) {                            vector<int> vt2;                            vt2.push_back(num[i]);                            vt2.push_back(num[j]);                            vt2.push_back(num[(*it2)[0]]);                            vt2.push_back(num[(*it2)[1]]);                            sort(vt2.begin(), vt2.end());                            mset.insert(vt2);                        }                    }                }                (hm[num[i]+num[j]]).push_back(v);            }        }        for (auto it3=mset.begin(); it3!=mset.end(); it3++) {            rt.push_back(*it3);        }        return rt;    }};

注意与总结：

①C++ 用了STL不知道 复杂度该怎么分析了

②避免重复要小心，

一种是排序后while判断前后是否重复

while (i + 1 < numsSize && nums[i + 1] == nums[i]){i++;}

一种使用set，set.insert() 重复的不会加入，map也是

一种使用 

auto tmp = unique(ret.begin(), ret.end());ret.erase(tmp,ret.end());

③ qsort用在内置类型的排序，sort(iterator.begin(),iterator.end(),cmp)一般用在有迭代器的

④指针、迭代器的->,对象的.，经常搞混，如果没有编译器提醒，直接写纸上的话会死很惨